Chemistry Thermochemistry

The specific heat of iron metal is .450 J/g-K. How many J of heat are necessary to raise the temperature of a 1.05 kg block of iron from 25C to 88.5C?

(.450) (1050) (63.5) = 30,000.75 J/g

How much heat energy, in Joules, is absorbed when 150 grams of water is warmed from 34.8C to 96.3C?

(150) (4.184) (61.5) = 38,597.4 J/g

Calculate the number of grams of Mg needed for this reaction to release enough energy to increase the temperature of 75 mL of water from 21C to 79C

(58) (75) (4.184) = 18.2 kJ (1 mol / -353 kJ) (24.31 / 1 mol) = 1.2 g

Glyceryl trioleate, the main ingredient in olive oil, burns with in oxygen gas according to the chemical equation below. When a 2.30 g sample of glyceryl trioleate is burned in a calorimeter that has a total heat capacity of 9.580 kJ/C, the temperature of the calorimeter increases from 23.201 C to 23.725C - C57H104O6 + 80 O2 -----> 57 CO2 + 52 H2O. Calculate the amount of heat absorbed by the calorimeter.

(SH) (Change in T) (9.58) (23.725 - 23.201) = 5.020 kJ

What would be the final temp of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter?

-(121) (.385) (Delta T) = (150) (4.184) (Delta T) -46.585 (Delta T) +4,677.13 = 627 (Delta T) -15,737.7 T = 5.22 25.1 + 5.22 = 30.3

Calculate the mass of CO2 produced per kJ of heat emitted

-1035.2 / 2 mol (2 mol / 44g) = 23.52 kJ/g ---> 1/23.52 g/kJ

Calculate the heat produced by the by combustion per liter of methanol. Methanol has a density of .791 g/mL.

-1035.2/2 (1 mol / 32g) (.791 g / 1 mL) (1000 mL / 1 L = 2.5 x 10^4 kJ/L

Calculate the heat change for the formation of nitrogen monoxide from its elements.

-1170 / 2 = -585 kJ 1530 / 2 = 765 765 + -585 = 180 kJ

Calculate the heat produced per liter of ethanol by combustion of ethanol under constant pressure. Ethanol has a density of .789 g/mol

-1234.7 kJ / 1 mol (1 mol / 46 g) (.789 g / 1 mL) (1000 mL / 1 Liter) = 21,177 kJ/L

Calculate the mass of CO2 produced per kJ of heat emitted

-1234.7 kJ/ 2 mol (1 mol / 44 g) = 14 kJ/g ----> 1/14 g/kJ

Calculate the delta H for the reaction of 2NO + O2 -----> 2NO2.

-183 + 33(2) = -117 kJ

Calculate the standard enthalpy change for the reaction assuming H2O (g) as a product. C2H5OH - -277.7 kJ O2 - 0 H2O - -241.8 kJ CO2 - -393.5 kJ

-277.7 ----> 3(-241.8) + 3(-393.5) -277.7 -----> -1512.4 -1512.4 - -277.7 ----> -1234.7 kJ/mol

What volume of SO3 would be produced if 50J of heat is transferred at 20C and 1 atm?

-50 kJ (2 mol / -198.2) = .5045 moles PV=nRT -------> (1) (v) = (.5045) (293) (.0821) = 12.14 L

How many grams of magnesium oxide are produced during an enthalpy change of -96 kJ?

-96 kJ ( 2 mol / -1204 kJ) (403 g / 1 mol) = 6.02 grams

Two solutions, initially at 24.60C, are mixed in a coffee cup calorimeter. The heat capacity of the calorimeter is 15.5 J/C. When a 100 mL volume of .100 M silver nitrate is mixed with a 100 mL sample of .200 M sodium chloride solution, the temperature in the calorimeter rises to 25.30C. Determine the delta H in kJ/mole.

.1 L (.1 mol / 1 L) = .01 mol AgNO3 .1 L (.2 mol / 1 L ) = .02 mol NaCl 200 (.7) (4.184) = 585.76 15.5 (.7) = 10.85 585.76 + 10.85 = -596.61 kJ -596.61 / .01 = -59.7 kJ / mol

What is the percent yield of the SiC equation?

.83 moles 100 = x x = 96.34% 33.425 32.2

MREs are military meals that can be heated on a flameless heater. The heat produced by the following reaction: Mg + 2H2O ---> Mg(OH)2 + H2. Calculate the standard enthalpy reaction Mg=0 H2O = 2(-285.83) Mg(OH)2 = -924.7 H2 = 0

0 + 2(-285.83) ---> -924.7 + 0 -927.7 + 2(285.33) = -353 kJ

Consider the following reaction: 1 HCH3COO- + 2O2 ----> 2H2O + 2CO2. Delta H = -871 kJ. Draw a complete enthalpy diagram for this reaction.

1 HCH3COO- + 2O2 \ \ \ -871 kJ \ \/ 2H2O + 2CO2

The complete combustion of acetic acid releases 871 kJ of heat per mole of the acid. Write a balanced thermochemical equation for the reaction/

1 HCH3COO- + 2O2 ----> 2H2O + 2CO2

When 50 grams of silicon dioxide is heated with an excess of carbon, 32.2 grams of SiC is produced. Write the balanced equation.

1 SiO2 + 3C ---> 1SiC + 2CO

How many grams of ammonium nitrate must be dissolved in water so that 108 kJ of heat is released from the water?

1 mol (80 grams / 1 mol) = 80 grams 80 = 500 x = 517.5 kJ 82.8 x

How many liters of nitrogen gas form if 100 grams of hydrazine reacts with 100 grams of oxygen gas?

1 mol N2H4 (32 g / 1 mol) = 32 grams; 100 / 32 = 3.125 mol 1 mol O2 (32 / 1 mol) = 32 grams; 100 / 32 = 3.125 3.125 mol (22.4) = 70 L

How much heat in kJ is released when 50 grams of sodium hydroxide is dissolved in water?

1 mol NaOH (40 grams / 1 mol) = 40 grams 40 grams = 50 x = -556.25 kJ 445 x

A house is designed to have passive solar energy features. Brickwork incorporated into the interior of the house acts as an absorber. Each brick weighs about 1.8 kg. The specific heat of the brick is .85 J/g-K. How many bricks must be incorporated into the interior of the house to provide the same total heat capacity as 1.7 x 10^3 gals of watee

1.8 kg (1000 g/1 kg) = 1800 g (.85 J / 1 g) = 1530 J/brick 1700 gal (3.785 L / 1 gal) = 6426 (1000 mL / 1 L) = 6.4 x 10^6 6.4 x 10^6 (4.18/1) = 2.6 x 10^7 J/mL 2.6 x 10^7 = 1530 x x = 1.74 x 10^4 bricks

What is the enthalpy change during the process in which 100 grams of water at 50C is cooled to ice at -30C. The specific heat of liquid water is 4.184 J/g-K. The specific heat of ice is 2.09 J/g-K. The molar heat of fusion for water is 6.01 kJ/mol.

100g (4.184) (50) / (1000) = -20.95 kJ 100g (1 mol / 18 g) (6.01 / 1 mol) = -33.39 kJ 100g (2.09) (30) / (1000) = -6.27 kJ -20.95 + -33.39 + -6.27 = -60.58 kJ

How much heat energy is needed to change a lock of ice at -20C to steam at 155C? The block has a dimension of 15 in per side.

15 in (2.54 cm / 1 in) = 38.1 cm^3 per side ---> 55306 (1 mL / 1cm) (.96 g / 1 mL) = 50,882 grams of ice H = (mass) (SH) (Delta T) -20 to 0 = (50,882) (20) (2.09) = 212681 J 0 to 100 = (50,882) (100) (4.184) = 21289029 J 100 to 150 (50,882) (50) (1.84) = 5149258 J Plateau at 0 = 50,882 (1 mol / 18 g) (6.01 / mol) = 16989 kJ Plateau at 100 = 50,882 (1 mol / 18 g) (40.1 / mol) = 115050 kJ 2127 + 21289 + 16989 + 5149 + 115050 = 160,604 kJ

Calculate the heat change for the formation of P4O10.

1640.1 + -2940.1 = 1300 kJ

Consider the following reaction: 2Mg + O2 ---> 2MgO. Delta H = -1204 kJ. Draw a complete enthalpy diagram for this reaction.

2 Mg + O2 \ \ \ -1204 kJ \ \/ 2MgO

When magnesium metal combines with oxygen from the air, 1204 kJ of heat energy is released. Write the balanced thermochemical equation.

2 Mg + O2 -----> 2MgO

What amount of heat is produced if 2.075 grams of SO3 are made?

2 mol SO3 (80 g / 1 mol) = 160 g 160 g 2.075 -198.2 x x = -2.6 kJ

Calculate the standard enthalpy change for the reaction assuming H2O g as a product CH3OH - -238.6 kJ O2 - 0 H2O - -241.8 kJ CO2 - -393.5 kJ

2(-238.6) + 0 ----> 4(-241.8) + 2(-393.5) -477 -----> -1512.4 -1512.4 - -477 ----> -1035.2 kJ/mol

In the reaction 2SO2 + O2 ---> 2SO3, calculate Delta H if: SiO2 = -296.1 kJ O2 = 0 SiO3 = -395.2 kJ

2(-296.1) + 0 ---> 2(-395.2) -592.2 -----> -790.4 -790.4 + 592.2 = -198.4

Calculate the enthalpy change for the combustion of one mole of glyceryl trioleate.

2.3 g (1 mol /886 g) = 2.59 x 10^-3 -5.020 / 2.59 x 10^-3 = -1.94 x 10^3

Calculate the enthalpy change for the combustion of one mile of glyceryl trioleate.

2.3 g (1 mol/ 886 g) = 2.59 x 10^-3 -5.020 / 2.59 x 10^-3 = -1.94 x 10^3

Calculate the amount of heat transferred when 2.4 g of Magnesium reacts.

2.4 grams (1/48 g) .05 = 1 x = -60.2 kJ x -1204

Calculate the values of final energy per empirical formula unit of the hydrocarbon

21.83 grams (1 mol / 44 g) (-393.5 kJ / 1 mol) = -195.2 kJ 4.47 grams (1 mol / 18 g) (-285.83 / 1 mol) = -70.74 kJ -195.2 + -70.94 + 311 = -44.9 (2) = -89.72

A sample of hydrocarbon is combusted completely in O2 to produce 21.83 g CO2, 4.47 g H2O and 311 kJ of heat. What is the mass of the hydrocarbon sample that was combusted?

21.85 g (12g / 44g) = 5.959 g 4.47 g (1 mol / 18g) (2 mol / 1 mol) = .5001 5.959 + .5001 = 6.4591 grams

Use the thermochemical equation for the combustion of methane written below to answer the following question - CH4 + 2O2 ---> CO2 + 2H2O where Delta H = -802.3 kJ/mol. How much heat is released when 24.8 grams of CH4 is burned in excess oxygen gas to produce carbon dioxide and water?

24.8 g ( 1 mol / 16 g) (-802.3 / 1 mol) = -1210 kJ

Use the thermochemical equation for the combustion of methane written below to answer the following question - CH4 + 2O2 -----> CO2 + 2H2O - Delta H = -802.3 kJ/mol. How much heat is released when 24.8 g of CH4 is burned in excess oxygen gas to produce carbon dioxide and water?

24.8 g (1 mole / 16 g) (-802.3 kJ / 1 mole) = -1210 kJ

1674 J of heat are absorbed by 25 mL of an aqueous solution of NaOH. The density of this solution is 1.10 g/mL and the specific heat of the solution is 4.10 J/gC. The temperature of the solution goes up how many degrees?

25 mL (1.10 g / 1 mL) = 27.5 grams 1674 = (27.5) (4.10) (x) x = 14.8 C

14.0 grams of metal at 24C has 250 joules of heat added to it. The metal has a specific heat of .105 J/gC. What is the final temperature of the metal?

250 J = (.105) (14) (x-24) x = 194C

Consider the following reaction: 2Al2O3 ----> 4Al + 3O2. Delta H = 3352kJ. Draw a complete enthalpy diagram for this reaction.

2Al2O3 /\ \ \ 3352 kJ \ 4Al + 3O2

Exactly 3352 kJ of heat is required for the decomposition of aluminum oxide into its elements. Write a balanced thermochemical equation for the reaction.

2Al2O3 ----> 4Al + 3O2

CH3OH is used as a fuel in race cars. Write a balanced equation for the combustion of liquid CH3OH in air

2CH3OH + 3O2 ----> 4H2O + 2CO2

Which has greater enthalpy, 3 moles of C2H2 or 1 mole of C6H6?

3 moles of C2H2

Use the information below to determine the heat reaction for 3Fe2O3 + CO ---> 2Fe3O4 + CO2 Fe2O3 - -824 kJ/mol Fe3O4 - -1118 kJ/mol CO - -111 kJ/mol CO2 - -394 kJ/mol

3(-824) + -111 ----> 2(-1118) + -394 -2583 -------> -2630 -2630 - - 2583 = -47 kJ/mol

When power was turned off to a 30 gallon water heater, the temperature of the water dropped from 75C to 22.5C. How much heat was lost to the surrounding in kilojoules?

30 gallon (3.785 Liters/ 1) = 113.55 113.55 (52.5) = 5,961.38 kJ (4.184) = 24,942.41 kJ

How many grams of CO gas are made in 1 SiO2 + 3C ---> 1SiC + 2CO?

32.2 / 40.11 = .80279 (2/1) (28/1) = 44.95 g

The difference between your answers for a and b is due to the heat lost through the Styrofoam cups and the heat necessary to raise the temperature of the inner wall of the apparatus. The heat capacity of the calorimeter is the amount necessary to raise the temperature of the apparatus by 1 K. Calculate the heat capacity of the calorimeter in J/K.

3275 - 3135 / (30.1 - 25.1) = 23.0 J/K

The hydrocarbon C2H2 and C6H6 have the same empirical formula. Benzene is an aromatic hydrocarbon, one that i usually stable because of its structure. By using the data in appendix C, determine the standard enthalpy change for 3C2H2 ----> C6H6

3C2H2 ----> C6H6 3(226.77) ----> 49 49 - 3(226.77) = -631.3 kJ

Calcium carbonate reacts with phosphoric acid to produce calcium phosphate, carbon dioxide and water. What is the balanced equation?

3CaCo3 + 2H3PO4 ---> Ca3(Po4)2 + 3CO2 + 3H2O

Consider 6O2 + 6H2O ----> C6H12O6 + 6O2 H = 43 kJ If 4.00 kJ of heat are added, how many liters of gas can be produced at STP?

4 kJ / 1 mol (6 mol / 43 kJ) (22.4 L / 1 mol) = 12.5 L

A chunk of silver has a heat capacity of 42.8J/C. If the silver has a mass of 181 grams, calculate the specific heat of silver.

42.8 / 181 = .236 J/g-C

When a certain substance with a mass of 100 grams is heated from 25C to 75C, it absorbed 4500 Joules of heat energy. Calculate the specific heat of the substance and identify it using the following table: Water: 4.184 J/g-K Ice: 2.1 J/g-K Aluminum: 0.90 J/g-K Silver: 0.24 J/g-K Mercury: 0.14 J/g-K

4500 = (100) (SH) (50) SH = .9 J/g-K Aluminum

How do you make these equations fits the target equation: 4NH3 + 3O2 ---> 2N2 + 6H2O H = -1530 kJ 4NH3 + 5O2 ---> 4NO + 6H2O H = -1170 kJ

4NH3 + 3O2 ---> 2N2 + 6H2O - Flip and divide by two 4NH3 + 5O2 ---> 4NO + 6H2O - Divide by two

How much heat is needed to raise the temperature of 5.28 gallons of water from 25C to 88C?

5.28 gallons (3.785 L / 1 gal) = 19.9848 L 19.9848 L (63 degree / 1) = 1,259.0424 kJ 1,259.0424 kJ (4.184 kJ / 1) = 5,267.8 kJ

What is the empirical formula of the hydrocarbon?

5.959 (1 mol / 12.4 g) = .496 mol .5001 (1 mol / 1.008 g) = .496 CH

How many molecules of carbon dioxide form when 50 grams of calcium carbonate reacts with 70 grams of phosphoric acid?

50 / 100 = .5 mol 70 / 98 = .71 mol .5 (6.02 x 10^23) = 3.01 x 10^23 molecules

How much heat in kJ is absorbed when 50 grams of liquid water at 75C is converted to steam at 120C? The molar heat of vaporization for water is 40.7 kJ/mole. The specific heat of water is 4.184 J/g-K. The specific heat of steam is 1.84 J/g-K.

50g (1 mol / 18.015) (40.7 grams / 1 mol) = 112.96 kJ 50g (4.184) (25) = 5230 J 50g (1.84) (20) = 1840 J 112.96 kJ + 5.23 kJ + 1.84 kJ = 120.03 kJ

How much heat is released when 50 grams of ethanol completely combusts?

50g (1/46 g) = 1.0869 1.0869 = 1 x = -1342.32 kJ x 1235

How many kilojoules of heat are absorbed when 7.5 grams of magnesium oxide are decomposed?

7.5 grams (1 mol / 40.3 grams) (602 grams / 1 mol) = -112.875 kJ

Calculate the value of delta H for N2O4 + 4H2 ---> N2 + 4H2O if: N2O4 - 9.16 kJ H2 - 0 N2 - 0 H2O - -241.8 kJ

9.16 + 0 = 0 + 4(-241.8) -967.20 - 9.16 ---> -978.3 kJ

Calculate the standard enthalpy of formation of gaseous B2H6 using the following thermochemical equation Target equation: 2B + 3H2 ---> B2H6 B4 + 3O2 ---> 2B2O3 h = -2509.1 kJ 2H2 + O2 ---> 2H2O h = -571.7 kJ B2H6 + 3O2 ---> B2O3 + 3H2O h = -2147.5 kJ

B4 + 3O2 ---> 2B2O3 - Divide by 2 2H2 + O2 ---> 2H2O - Divide by 2/3 B2H6 + 3O2 ---> B2O3 + 3H2O - Leave -2509.1 / 2 = -1254.55 -571.7 / 2/3 = -857.55 -2145.7 =35.4 J

Determine the value of the standard enthalpy change for the following reaction using the information below - Target equation: 2Fe2O3 + 3C ---> 4Fe + 3CO2 C + O2 ----> CO2 Delta H = -393.5 kJ/mol 8Fe + 6O2 ----> 4Fe2 Delta H = -3296.8 kJ/mol

C + O2 ----> CO2 - Multiply by 3 8Fe + 6O2 ----> 4Fe2 - Flip and divide by 2 3 (-393.5) ----> 3296.8 (1/2) -1180.5 --------> 1648.4 Delta T = 467.9 kJ

From the following data for three prospective fuels, calculate which could provide you with the most energy per unit volume.

C2H5NO2 - (1.052 g / 1cm^3) (1 mol / 75 g) (1,368) = -19.17 kJ C2H5OH - (.789 / 1 cm^3) (1 mol / 46 g) (1367) = -23.96 kJ CH6N2 - (.874 / 1cm^3) (1 mol / 46) (1307) = 24.83 kJ CH6N2

C2H5OH is currently blended with gasoline as an automobile fuels. Write a balanced equation for the combustion of liquid ethanol in air.

C2H5OH + 3O2 -----> 3H2O + 2CO2

Gasohol contains ethanol, C2H5OH, which completely combusts to produce 1235 kJ of heat energy. Write the balanced thermochemical equation.

C2H5OH + 3O2 ----> 2CO2 + 3H2O

Write the balanced thermochemical equation of the complete combustion of 1 mole of C3H6O that liberates 1790 kJ.

C3H6O + 4O2 ---> 3CO2 + 3H2O

Determine the value of the standard enthalpy change for the following reaction using the information below - Target equation: H2O l ----> H2O w CH4 + 2O2 ----> CO2 + 2H2O l Delta H = -890.5 kJ/mol CH4 + 2O2 ----> CO2 + 2H2O g Delta H = -802.3 kJ/mol

CH4 + 2O2 ----> CO2 + 2H2O l - Divide by 2 CH4 + 2O2 ----> CO2 + 2H2O g - Divide by 2 -401.15 - - 445.25 ----> 44.1 kJ/mol

A coffee cup calorimeter contains 150 grams of water at 25 JC. A 121 g block of copper metal is heated to 100.4C by putting it in a beaker of boiling water. The specific heat of Cu is .385 J/g-K. The Cu is added to the calorimeter and after a time the contents of the cup reach a constant temperature of 31C. Determine the amount of heat in J lost by the copper block.

Cu- (121) (.385) (100.4 - 30.1) = -3275 kJ

Is 2SiO2 + O2 ---> 2SiO3 exo or endo?

Exo

Consider the following reaction: 2Mg + O2 ---> 2MgO. Delta H = -1204 kJ. Is this reaction exothermic or endothermic? Explain your choice.

Exothermic because the value is negative. The product lost energy.

A pound of plain M and Candies contain 96 g fat, 320 g carbohydrates, and 21 g protein. What is the fuel value in kJ in a 42 gram serving? How many calories does it provide?

Fat = (.0925) (96) (38) = 337.85 kJ Carbs and Proteins = (.0925) (341) (17) = 536.771 kJ 337.85 + 536.771 = 870 870 / 4.184 = 210 calories

Which is a greater source of energy, 5g of fat or 9g of carbohydrates Fat - 38 per g Carb - 17 per g

Fat = 5(38) = 190 kJ/g Carb = 9(17) = 153 kJ/g Fat is a greater source of energy

Find the amount of heat lost of gained by the calorimeter in Zn (s) + 2 HCl (aq) --> ZnCl2(aq) + H2 (g).

Formula of heat change, ∆T = T2 - T1 = 29.02 - 20.51 = 8.51 C Specific heat capacity for water C = 4.184 J/g-C Mass of water = 1.50 x 102 ml = 150 g (for water 1 ml = 1 g ) Use formula of enthalpy change ∆H = m * C *∆T Plug the values ∆H = 150 g x 4.184 J/g-C x 8.51 C ∆H = 5341 Joules Divide by 1000 to convert in kJ = ∆H = 5.34 KJ Temperature is increasing so that it is a exothermic reaction And for all exothermic reaction ∆H is always negative energy lost by reaction and gained by solution. hence calorimeter gain this energy So final answer will ∆H = - 5.34 KJ

50.0 g of iron has an initial temperature of 225C. This iron is bought into contact with 50.0 g of gold that has an initial temperature of 25C. Assuming that no heat is lost to the surroundings, what will be the final temperature when the two metals reach thermal equilibrium. The specific heat of gold in .128 J/gC and the specific heat of iron is .449 J/gC.

H gained Au = Heat lost Fe (mass Au) (delta Temp Au) (Sh Au) = (mass Fe) (delta Temp Fe) (Sh Fe) (50.0) (x-25) (.128) = (50.0) (225-x) (.449) x = 180.63C

A coffee cup calorimeter contains 150 grams of water at 25 JC. A 121 g block of copper metal is heated to 100.4C by putting it in a beaker of boiling water. The specific heat of Cu is .385 J/g-K. The Cu is added to the calorimeter and after a time the contents of the cup reach a constant temperature of 31C. Determine the amount of heat in J gained by the water if the specific heat of water is 4.184 J/g-K

H2O - (150) (4.184) (30.1-25.1) = 3135 kJ

Glyceryl trioleate, the mean ingredient in olive oil, burns with in oxygen gas according to the chemical equation below. When a 2.30 g sample of glyceryl trioleate is burned in a calorimeter that has a total heat capacity of 9.580 kJ/C, the temperature of the calorimeter increases from 23.201 C to 23.725 C: C57H104O6 + 80 O2 -----> 57 CO2 + 52 H2O. Calculate the amount of heat absorbed by the calorimeter.

Heat absorbed = (HC) (Delta T) (9.58)(23.726 - 23.201) = 5.020 kJ

A student mixed 75mL of water containing 0.50 mol HCL at 22.5C with 75mL of water containing NaOH at the same temperature in a foam cup calorimeter. The temperature of the resulting solution increased to 26C. How much heat in kJs was released by this reaction? Assume the density of the resulting solution was 1.0 g/mL.

J = (150) (4.184) (3.5) kJ = (2196.6) / (100) = -2.1966 kJ

If the target equation is 2NO + O2 -----> 2NO2, what must you do to get the target equation using the following equations: N2 + O2 ----> 2NO H = 183 kJ 1/2 N2 + O2 -----> NO2 H = 33 kJ

N2 + O2 ----> 2NO - Flip 1/2 N2 + O2 -----> NO2 - Multiply by 2

What is the target balanced equation for the formation of nitrogen monoxide from its elements?

N2 + O2 ---> 2NO

N2H4 is used as rocket fuel. It reacts with oxygen to form nitrogen gas and water vapor. Write the balanced equation.

N2H4 + O2 ---> N2 + 2H2O

When ammonium nitrate is dissolved into water, ammonium and nitrate ions are released absorbing 82.8 kJ of heat energy from the water. Write this balanced thermochemical equation.

NH4NO3 ----> NH4+ + NO3-

When solid sodium hydroxide is dissolved into water, forming aqueous sodium ions and hydroxide ions, 445 kJ/mol of heat energy is released. Write the balanced thermochemical equation for this physical process.

NaOH ---> Na+ + OH-

Find the limiting reactant of Zn (s) + 2 HCl (aq) --> ZnCl2(aq) + H2 (g).

Number of moles of Zn = given mass of Zn / molar mass of Zn 1.75 g Zn / 65.3.9 g/mol = 0.02676 moles Zn In equation 1 mole of Zn react with 2 mol of HCl Hence 0.02676 moles Zn react with = 2 x 0.02676 = 0.05352 mol of HCl But required moles of HCl (0.05352 mol) > given moles (0.0375) Hence HCl is limiting reagent

If the target equation is P4O6 + 2O2 ---> P4O10, how can we make the following equations fits the target equation: P4 + 3O2 ---> P4O6 H = -1640.1 P4 + 5O2 ---> P4O10 H = -2940.1

P4 + 3O2 ---> P4O6 - Flip P4 + 5O2 ---> P4O10 - Keep

Heating an ore of Sb2S3 in the presence or iron given the element antimony and iron II sulfide. Write the balanced equation.

SB2S3 + 3Fe ---> 2Sb + 3FeS

What is fuel value?

The energy released when one gram of any substance is combusted

How does the human body expel these reaction products?

The human body expels CO2 and H2O by going to the bathroom

A coffee cup calorimeter contains .0375 mol HCl dissolved in 1.50 x 10^2 mL of water at 20.51 C. When 1.75 grams of solid zinc metal is placed in the calorimeter the temperature rises to 29.02C. Assume that no heat is lost to the surroundings. Write a balanced equation for the reaction that takes place in the calorimeter.

Zn (s) + 2 HCl (aq) --> ZnCl2(aq) + H2 (g)

A coffee cup calorimeter contains .0375 mol HCl dissolved in 1.50 x 10^2 of water at 20.51C. When 1.75 of solid zinc metal is placed in the calorimeter the temperature rises to 29.02 C. Assume no heat is lost to the surrounding. Write the balanced equation for the reaction that takes place in the calorimeter.

Zn + 2HCl -----> ZnCl2 + H2

Find the limiting reagent in the equation Zn + 2HCl -----> ZnCl2 + H2

Zn = 1.75 / 63.4 = .02676 moles * 2 = .05352 moles HCl = .0375 HCl is limiting

Using the formula, calculate the enthalpy of the formation of acetone.

x + 0 = 3(-393.5) + 3(-241.8) 1905.90 - x = -1790 kJ x = 116.5 kJ