Question

solve problem E-L

1. Predict product(s) and write a balanced chemical reaction for each problem. If there is only one product. It is otherwise there will be two products. 2. Find the limiting reactant. 3. Calculate the mass of the indicated product. Find mass of indicated product Reactant #1 Reactant #2 A B 1.067g nitrogen gas 6.97g of aluminum metal 25.0mL of 0.167M sodium phosphate solution nitrogen monoxide gas (only one product) solid aluminum nitride (only one product) find mass of the solid product in this precipitation reaction. D E 6.20 g oxygen gas 6.97 g carbon dioxide gas 2.0678 oxygen gas 6.97 g of nitrogen gas 25.0 mL of 0.250M magnesium chloride solution 1.462 g solid sulfur 8.69 g of solid calcium oxide 1.467 g of solid iron (III) oxide 35.0 mL of 0.280M barium chloride solution sulfur trioxide (only one product) solid calcium carbonate (only one product) 6.02 g solid carbon carbon dioxide gas and solid iron metal. Find mass of solid iron metal. find mass of the aqueous product in this precipitation reaction. 35.0 mL of 0.167M potassium sulfate solution 12.4 g propane gas (C3H8) 24.02 oxygen gas 2.34 g zinc metal 50.0 mL of 0.15M copper (11) chloride solution 23.4 g bromine liquid 125.5 mL of 0.25M barium iodide solution 155.5 mL of 1.35M silver 2.06 g magnesium metal () nitrate solution 8.401 g manganese metal 20.3 g nitrogen gas carbon dioxide gas and water vapor. Find mass of water vapor. find mass of copper metal in single displacement reaction. find mass of barium bromide in single displacement reaction. find mass of silver metal in single displacement reaction. solid manganese (V) nitride K L

Answer 1

Reactant 1 moles (mass/molar mass)	Reactant 2 moles	Balance equation	Limiting reactant (one with low number of moles)	Mass of product
Carbon dioxide gas moles = 6.97 g/44.01 g/mol = 0.158	Calcium oxide = 8.69 g/56.08 = 0.155	CaO + CO2 ----------------> CaCO3	Calcium oxide	Molar ratio of calcium oxide and product = 1:1 Moles of product = 0.155 Mass of product = moles x molar mass = 100.08 g/mol x 0.155 mol = 15.5 g
Solid carbon = 6.02 g/12.01 g/mol = 0.501	Solid iron oxide = 1.467 g/159.69 g/mol = 0.0092	Fe2O3 + 3C = 2Fe(s) + 3CO2	1 mole of Fe2O3 require 3 moles of C 0.0092 moles --> 3 x 0.0092 = 0.0276 moles But C is in excess. Hence solid iron oxide is limiting reactant	Molar ratio of solid iron oxide and solid iron = 1:2 Moles of solid iron = 0.0092 x 2 = 0.0184 moles Mass of solid iron = 0.0184 mol x 55.845 g/mol = 1.03 g
K2SO4 = Molarity x vol in lts = 0.167 M x 35/1000 = 0.005845	BaCl2 = 0.280 M x 35/1000 = 0.0098	BaCl2 + K2SO4 = 2KCl(aq) + BaSO4(s)	Molar ratio of BaCl2 and K2SO4 = 1:1 Hence limiting reactant is K2SO4	Mass of aqueous product (KCl) : Molar ratio of K2SO4 and KCl = 1:2 Hence moles of KCl = 0.005845 x 2 = 0.01097 moles Mass of KCl = 74.55 x 0.01097 = 0.817 g
Propane gas = 12.4 g/44.1 g/mol = 0.281 moles	oxygen = 24.2 g/32 g/mol = 0.751 mol	C3H8 + 5O2 ---> 3CO2 + 4H2O	Molar ratio of propane and oxygen = 1 :5 0.281 moles of propane require 5 x0.281 = 1.405 moles of oxygen But given oxygen gas molecules is 0.751. Hence limiting reactant is oxygen	Molar ratio of oxygen and water vapor = 5:4 5 moles of oxygen produce 4 moles of water vapor 0.751 moles of oxygen produce 0.6008 moles of water vapor Mass of water vapor = 0.6008 mol x 18.01 g/mol = 10.82 g
Zinc metal = 2.34 g/65.38 g/mol = 0.0358	CuCl2 = M x V = 0.15 M x 50/1000 = 0.0075	Zn + CuCl2 = ZnCl2 + Cu(s)	Molar ratio of zinc and cuprous chloride = 1:1 Hence moles of cuprous chloride is limiting reactant	Molar ratio of CuCl2 and Cu = 1:1 Moles of Cu = 0.0075 mol Mass of Cu = 0.0075 mol x 63.55 g/mol = 0.477 g
Bromine liquid = 23.4/159.1 = 0.147	BaI2 = 0.25 M x 125.5/1000 = 0.0314	BaI2 + Br2 --> BaBr2 + I2	Molar ratio of BaI2 and Br2 = 1:1 Hence BaI2 is limiting reactant	Molar ratio of BaI2 and BaBr2 = 1:1 Moles of BaBr2 = 0.0314 Mass of BaBr2 = 0.0314 x 297.14 g/mol = 9.33 g
AgNO3 = 1.35 M x 155.5/1000 = 0.2099	Mg = 2.06 g/24.31 g/mol = 0.0847	2AgNO3 + Mg ---> Mg(NO3)2 + 2Ag	Molar ratio of silver nitrate and magnesium = 2:1 2 moles of silver nitrate require 1 mole of Mg 0.2099 moles --> 0.10495 moles of Mg But moles of Mg is less, hence it is limiting reactant	Molar ratio of Mg and Ag = 1:2 Hence moles of Ag = 0.0847/2 = 0.04235 Mass of Ag = 0.04235 x 107.87 = 4.568 g
Manganese = 8.401 g/54.94 g/mol = 0.153	Nitrogen gas = 20.3 g/28 g/mol = 0.725	6Mn+ 5N2 --->2Mn3N5	6 moles of Mn require 5 moles of nitrogen 0.153 --> 0.1275 moles of nitrogen But given nitrogen gas is in excess hence manganese is limiting reactant	Molar ratio of Mn and Mn3N5 = 6:2 6 moles of Mn produce 2 moles of product 0.153 moles --> 0.051 moles of product Mass of Mn3N5 = 0.051 mol x 234.85 g/mol = 11.98 g