Chemistry Chapter 6

Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, according to the equation 3H2(g)+N2(g)?2NH3(g) Part A How many moles of NH3 can be produced from 18.0mol of H2 and excess N2? Express your answer numerically in moles.

# mol NH3 = 18.0 mol H2 x 2 mol NH3 3 mole H2 =12 mol NH3

If you had excess aluminum, how many moles of aluminum chloride could be produced from 28.0 g of chlorine gas, Cl2?

1) If we had excess Cl2, then limiting reagent is Al so moles of Al will be the same as the moles of AlCl3 produced at the output moles of Al = 23/27 = 0.8518 moles 2)If we had excess Al , then limiting reagent is Cl2 so moles of Cl2 will be the 3/2 times as the moles of AlCl3 produced at the output moles of Cl2 = 28/71 = 0.394 moles hence moles of AlCl3 = 0.394*2/3 = 0.2629 moles

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)---->2AlCl3(s) You are given 23.0 g of aluminum and 28.0 g of chlorine gas. What is the maximum mass of aluminum chloride that can be formed when reacting 23.0 g of aluminum with 28. g of chlorine?

1) to judge which reactant is the limiting reagent. moles of Al = mass/atomic mass = 23.0/27 = 0.85 mol moles of Cl2 = 28.0/71 = 0.39 mol since the molar ratio of Al : Cl2 = 2 : 3, moles of Al that reacts with 0.39 mol of CL2 = 0.39x2/3 = 0.26 mol < 0.85 mol thus, Al is in excess and Cl2 is the limiting reagent. molar ratio of AlCL3 : Cl2 = 2 : 3 moles of AlCl3 = moles of Cl2 x 2 / 3 = 0.39 x 2/3 = 0.26 mol molar mass of AlCl3 = 133.3 g/mol mass of AlCl3 = moles x molar mass = 0.26 x 133.3 = 34.7 g

PART B:What mass of carbon dioxide is produced from the complete combustion of 5.10×10−3 g of methane? PART C: What mass of water is produced from the complete combustion of 5.10×10−3 g of methane? Express your answer with the appropriate units. Part D What mass of oxygen is needed for the complete combustion of 5.10×10−3 g of methane? Express your answer with the appropriate units.

1.40×10−2 g CH4(g)+ 2O2(g)------> CO2(g)+2H2O(g)from the equation , we can see , 1 mole methane produce s 1 mol CO2 and 2 mol H2O and will need 2 mol O2 so , we can say , 16 g of methane will produce 44 g CO2 and 36 g water and will need 64g O2 so , 5.10*10^-3 g methane produce CO2 = 5.1*10^-3 * (44/16) = 14.025*10^-3 g CO2 similarly , 5.10*10^-3 g methane produce H2O = 5.1*10^-3 * (36/16) =11.475*10^-3 g H2O similarly , O2 needed = 5.1*10^-3 * (64/16) = 20.4*10^-3 g O2 needed

How many moles of NH3 can be produced from 18.0 mol of H2 and excess N2?

12.0 mol NH3

Part A The balanced equation for the decomposition of water is shown below. 2 H2O → 2 H2 + O2 If 0.72 g of water react completely in this reaction, what is the theoretical yield of H2?

2 x moles of H2O = 2 x 0.72/18 = 0.0800 g/moles or 8.0×10−2 g

How many oxygen atoms are in 2.60 of quartz?

2.60 grams SiO2 (1 mole SiO2/60.09 grams)(2 mole O/1 mole SiO2)(6.022 X 10^23/1 mole O) = 5.21 X 10^22 atoms of quartz ( SiO2 is standard formulation, but SiO4 is actual tetrahedral structure of quartz, Change numbers if needed, but approach is the

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) You are given 23.0 g of aluminum and 28.0 g of chlorine gas. If you had excess chlorine, how many moles of of aluminum chloride could be produced from 23.0 g of aluminum?

2Al(s) + 3Cl2(g) ---> 2AlCl3(s) Always work in moles moles = mass of substane / molar mass 2Al (s) + 3Cl (g) --> 2AlCl3 (s) moles Al = 23.0 g / 26.98 g/mol = 0.852 moles Al Al is the limiting reactant

Determine the balanced chemical equation for this reaction. C8H18(g)+O2(g)→CO2(g)+H2O(g)

2C8H18(g)+25O2(g)→16CO2(g)+18H2O(g)

Notice that "SO4" appears in two different places in this chemical equation. SO 2?4 is a polyatomic ion called "sulfate." What number should be placed in front of CaSO4 to give the same total number of sulfate ions on each side of the equation? 3CaSO4 + 2AlCl3 --> 3CaCl2 + Al2(S04)3

Answer -> 3 is the number 3CaSO4 + 2AlCl3 --> 3CaCl2 + Al2(S04)3 How to solve: First, write out how many you have of each, like so: reactants: Ca: 1 SO: 4 Al: 1 Cl: 3 Products: Ca: 1 SO: 12 Al: 2 Cl: 2 Then, start with the hardest, which would be getting the same # of Cl on each side. 2&3 go into 6, so put a 3 in front of CaCl2 on the products side and a 2 in front of AlCl3 on the reactants side. What gives you 6 Cl on each side. Then, put a 3 in front of the CaSO4 on the products side, to get 12 SO. Now SO & Cl are balanced. So is Ca, & it became balanced on it's own from what you did with Cl. That leaves Al, just put a 2 in front of it on the products side & bam. Balanced. (:

How many grams of NH3 can be produced from 3.51 mol of N2 and excess H2.

N2 + 2H2 --> 2NH3 As you can see that the mole ratio is N2 : NH3 1 : 2 Hence, Mole of NH3 produced = 2 X 3.51= 7.02, In order to find grams, multiply it by its molecular mass. Mr of NH3 = 14 + 3 = 17 Mass of NH3 produced = 17 X 7.02 = 119.34 grams first you need the balanced equation N2 + H2 = NH3 we need it balanced N2 + 3H2 = 2NH3 now we have it balanced we have excess H2 so it doesnt play anything in the equation... 1 mole N2 produces 2 moles NH3 3.51 moles N2 ->> 2*3.51 moles NH3 we produce 7.02 moles of NH3 we need the molar mass of NH3 N = 14 grams H = 1 gram 14 + 3 = 17 grams per mole of NH3 mass produced = 7.02*17 = 119.34 g

Converting between quantities. To convert from a given quantity of one reactant or product to the quantity of another reactant or product:

first, convert the given quantity to moles. Use molar masses to convert masses to moles, and use Avogadro's number (6.02×1023 particles per mole) to convert number of particles to moles. Next, convert moles of the given reactant or product to moles of the desired reactant or product using the coefficients of the balanced chemical equation. For example, in the chemical equation 2H2+O2→2H2O the coefficients tell us that 2 mol of H2 reacts with 1 mol of O2 to produce 2 mol of H2O. Finally, convert moles of the desired reactant or product back to the desired units. Again, use molar masses to convert from moles to masses, and use Avogadro's number to convert from moles to number of particles.