Chemistry 1 Chemical Reactions A

Based on the reaction below, 36 g of HCl are mixed with excess Na2CO3. If the reaction is carried out at STP, how many liters of CO2 are formed? 2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g) (formula weight of HCl = 36.5) A. 5.6 L B. 11.2 L C. 22.4 L D. 44.8 L

(B) According to the stoichiometry of this reaction equation, 2 moles of hydrochloric acid react with one mole of sodium carbonate to form two moles of sodium chloride, one mole of water, and one mole of carbon dioxide. You are told in the question stem that there is an excess of sodium carbonate; therefore, the amount of carbon dioxide formed is dependent upon the amount of hydrochloric acid, the limiting reactant. Since there is 36 g of HCl, the number of moles of HCl is equal to 36 g/36.5 g/mol or approximately 1 mole. Since two hydrochloric acids form one carbon dioxide, the maximum amount of carbon dioxide that can form is 0.5 mol. At standard temperature and pressure, one mole of a gas occupies 22.4 L; therefore, 0.5 mol occupies 11.2 L—choice B is the correct response.

What is the correct expression for the equilibrium below? A. Kp= Pco Ph2/ Ph20 [C] B. Kp= Pco Ph2/ Ph20 C. Kp= PH20 [C] / Pco Ph2 D. Kp= Pco P^2 H2 / PH20

(B) For any gaseous equilibrium, the equilibrium constant (Kp) is equal to the partial pressures of the products raised to their stoichiometric coefficients, divided by the partial pressures of the reactants raised to their stoichiometric coefficients. Therefore, for the forward reaction, the equilibrium constant equals PCOPH2/PH2O. Remember that if the equilibrium involves pure solids or pure liquids, their concentrations do not change and they do not have to be included in the equilibrium expression. Since carbon is a solid, it is not included in the equilibrium expression and choice A can be eliminated. In the reaction expression, there is only one hydrogen, so the concentration of hydrogen in the equilibrium expression should not be raised to any exponent, and choice D can be eliminated.

Calculate the molar concentration of Ca2+ in a saturated solution of CaSO4 at 25C. (Ksp of CaSO4 = 1.4 10-5) A. 1.96 10-10M B. 3.7 10-3M C. 1.4 10-5M D. Cannot be determined

(B) In a saturated solution of calcium sulfate, the Ksp is equal to the product of the concentration of calcium ions and the concentration of sulfate ions: Ksp= 1.4 x 10 ^-5 = [Ca 2+]sat [SO4 2-]sat [Ca2+]sat[SO4 2-]sat = ([Ca2+]sat)^2 =1.4 x 10 ^-5 [Ca2+]sat = _______________ __________ \/ 1.4 x 10^-5 =\/14x10^-6= =4 x 10 ^-3 Since each unit of CaSO4 dissociates into one Ca2+ ion and one sulfate ion, the concentrations of the two ions are equal, i.e.: The only answer choice that is close to 4 × 10-3 is choice B.

What is the percent composition by weight of Al in Al2(SO4)3? (atomic weights: Al = 27, S = 32, O = 16) A. 8 % B. 16 % C. 32 % D. 48 %

(B)—16%. In one mole of Al2(SO4)3, there are two moles of aluminum, three moles of sulfur, and twelve moles of oxygen. This would correspond to 2 × 27 or 54 g of aluminum, 3 ×32 or 96 g of sulfur, and 12 × 16 or 192 g of oxygen. Therefore, the percent composition by weight of aluminum in aluminum sulfate is equal to the mass of aluminum in 2 mol of aluminum divided by the mass of 1 mole of aluminum sulfate (the molar mass) multiplied by 100. This is equal to 54/(54 + 96 + 192) × 100 or 54/342 × 100. You can approximate the answer by dividing 50 by 300 to give 1/6 = 0.1666..., and then multiplying by 100 to give about 16%—choice B.

4. In aluminum acetate (Al(C2H3O2)3), what element has the highest percentage by weight? (atomic weights: Al = 27, C = 12, H = 1, O = 16) A. A B. H C. O D. C

(C) In one mole of Al(C2H3O2)3, there is one mole of aluminum, six moles of carbon, nine moles of hydrogen, and six moles of oxygen. Let's translate the number of moles into actual weights. In one mole of Al(C2H3O2)3, there is 1 × 27 g (27 g) of aluminum, 6 × 12 g (72 g) of carbon, 9 × 1 g of hydrogen, and 6 × 16 g (96 g) of oxygen. Therefore, oxygen has the highest percentage by weight followed by carbon, aluminum, and finally hydrogen.

In the reaction 2 SO2 + O22 SO3, 0.25 mole of sulfur dioxide is mixed with 0.25 mole of oxygen and allowed to react. What is the maximum number of moles of SO3 that can be produced? A. 0.0625 moles B. 0.125 moles C. 0.25 moles D. 0.5 moles

(C) Initial quantities of both reactants are provided in this problem, so it is necessary to determine which reagent, if any, is in excess. The coefficients of the balanced equation, 2SO2 + O2 2SO3, indicate that two moles of SO2 are required for every mole of O2 in order to form two moles of SO3. Since the number of moles of reagent are equal, it follows that SO2 is the limiting reagent and O2 is in excess. Therefore, the maximum number of moles of SO3 that can be produced under these conditions is the same as the number of moles of SO2 that are consumed—0.25 moles. Therefore, choice C is the correct response.

What mass of H2SO4 contains 4 moles of hydrogen atoms? (atomic weights: H = 1, S = 32, O = 16) A. 49 g B. 98 g C. 196 g D. 392 g

(C) One mole of sulfuric acid contains two moles of hydrogens atoms, so two moles of sulfuric acid contains four moles of hydrogen atoms. To convert this into the mass of sulfuric acid, we have to multiply the number of moles of sulfuric acid by its molecular mass—2 mol × (2 + 32 + 16 × 4) g/mol = 2 mol × 96 g/mol = 196 g.

3. What is the empirical formula of a compound that contains 12.5 % hydrogen by mass, 37.5 % carbon by mass, and 50 % oxygen by mass? (atomic weights: C = 12, H = 1, O = 16) A. CH3O B. CHO C. CH4O D. CH2O

(C) The empirical formula is derived from the simplest integral ratio of elements in the compound. In order to calculate the empirical formula, imagine there is 100 g of the sample. According to the percentages, 37.5 g would be carbon, 12.5 g would be hydrogen, and 50 g would be oxygen. How many moles would each percentage correspond to? There would be 37.5/12 or approximately 3 mol of carbon, 12.5/1 or 12.5 mol of hydrogen, and approximately 3 mol of oxygen. Therefore, the ratio of moles of carbon to hydrogen to oxygen is approximately 3:12:3, or more simply, 1:4:1. Therefore, the empirical formula is CH4O, and choice C is the correct response.

What is the percent composition by weight of carbon in glucose (C6H12O6)? (atomic weights: C = 12, H = 1, O = 16) A. 72/ (72+12) * 100 B. 72/ (12+96) * 100 C. 96/ (72+12+96) *100 D. 72/ (72+12+96) *100

(D) The molecular formula of glucose is C6H12O6; therefore, in one mole of glucose there are six moles—or 6 × 12 g—of carbon. In terms of mass, we know that one mole of glucose has a mass of 6 × 12 g or 72 g (contributed by carbon), plus 12 × 1 g or 12 g (contributed by hydrogen), plus 6 × 16 g or 96 g (contributed by oxygen). Therefore, the percent composition by weight of carbon in glucose is equal to the mass of carbon in 1 mole of glucose divided by the mass of 1 mole of glucose (the molar mass) multiplied by 100. This is equal to 72/(72 + 12 + 96) × 100: choice D. Choices A and B can be eliminated because they incorrectly show the molar mass of glucose to be (72 + 12 g/mol) and (12 + 96g/mol), respectively. Choice C can be eliminated because this is the calculation for the percent composition by weight of oxygen in glucose, not carbon.

How many grams of oxygen are necessary to oxidize 88°grams of C3H8 to CO2 and H2O? (Atomic weights: C = 12, H = 1, O = 16) A. 80 g B. 120 g C. 160 g D. 320 g

(D) To answer this question correctly, you must first write and balance the equation for the combustion described: C3H8 + 5 O2→ 3 CO2 + 4 H2O Now you can see that for every mole of C3H8, 5 moles of oxygen will be required. The formula weight of propane is (3 × 12 + 8 × 1) = 44 g/mol, so 88 grams will be 2 moles and therefore 10 moles of oxygen will be required. At 32 g/mol, 10 moles of O2 is 320 grams.