Question

1. how many grams of aluminum oxide are produced according to the reaction below given that you start with 10.0 g of al and 19.0 grams of o₂? reaction: 4al + 3o₂ - 2al₂o₃ a) 5.00 b) 40.4 c) 18.9 d) 0.185 e) not enough information 25) 8.5 g sample of nh₃ on oxidation produces 4.5 g of no. calculate the percent yield. reaction: 4 nh₃ + 5 o₂ - 4 no + 6 h₂o a) 15 % b) 60% c) 30% d) 70% e) none of the above 26) which is the excess reagent in the following reaction given that you start with 15.5 g of na₂s and 12.1 g cuso₄? reaction: na₂s + cuso₄ - na₂so₄ + cus a) na₂so₄ b) cuso₄ c) na₂s d) cus e) not enough information 27) what is the theoretical yield in g of cus for the following reaction given that you start with 15.5 g of na₂s and 12.1 g cuso₄? reaction: na₂s + cuso₄ - na₂so₄ + cus a) 0.198 b) 0.0758 c) 18.93 d) 7.25 e) not enough information 28) how many grams of the excess reagent remain assuming the reaction goes to completion and that you start with 15.5 g of na₂s and 12.1 g cuso₄? reaction: na₂s + cuso₄ - na₂so₄ + cus a) 15.45 b) 9.58 c) 0.05 d) 5.92 e) not enough information

Explanation:

Step1: Calculate moles of reactants

Molar mass of \(Na_2S = 78.05\ g/mol\), moles of \(Na_2S=\frac{15.5\ g}{78.05\ g/mol}\approx0.199\ mol\)
Molar mass of \(CuSO_4 = 159.61\ g/mol\), moles of \(CuSO_4=\frac{12.1\ g}{159.61\ g/mol}\approx0.0758\ mol\)

Step2: Determine limiting reactant

From the reaction \(Na_2S + CuSO_4
ightarrow Na_2SO_4 + CuS\), mole ratio is \(1:1\). Since \(0.0758\ mol\) (\(CuSO_4\)) is less than \(0.199\ mol\) (\(Na_2S\)), \(CuSO_4\) is the limiting reactant.

Step3: Calculate moles of \(CuS\) produced

Mole ratio of \(CuSO_4\) to \(CuS\) is \(1:1\), so moles of \(CuS = 0.0758\ mol\)

Step4: Calculate mass of \(CuS\)

Molar mass of \(CuS = 95.61\ g/mol\), mass of \(CuS=0.0758\ mol\times95.61\ g/mol\approx7.25\ g\)

Answer:

D. 7.25