Question

2 Zn (s) + O₂ (g) → 2 ZnO (s)

If 25 g of O₂ reacts with excess Zn in the reaction above, how many grams of ZnO will be formed?

Question 2 options:

	1.6 g ZnO
	32 g ZnO
	64 g ZnO
	127 g ZnO

H₃PO₄ + 3 NaOH → Na₃PO₄ + 3 H₂O

If 3.3 moles of H₃PO₄ react with 6.2 moles of NaOH, how many moles of H₂O are formed? What is the limiting reactant?

Question 3 options:

	6.2 moles of H2O are formed; NaOH is the limiting reactant
	9.9 moles of H2O are formed; NaOH is the limiting reactant
	6.2 moles of H2O are formed; H3PO4 is the limiting reactant
	9.9 moles of H2O are formed; H3PO4 is the limiting reactant

2 H₂ + O₂ → 2 H₂O

What is the percent yield if 17 g of H₂O are formed after 6.5 g of H₂ reacts with 21 g of O₂?

Question 4 options:

	81%
	72%
	38%
	29%

Answer 1

2 Zn (s) +   O2 (g)   →     2 ZnO (s)

no of mole of O2 = 25/32 = 0.78 mole

no of mole of ZnO produced = 0.78*2 = 1.56 mole

mass of ZnO produced = 1.56*81.41 = 127 g

H3PO4   +    3 NaOH →   Na3PO4    +   3 H2O

1 mole H3PO4   =     3 mole NaOH

limiting reagent = NaOH

no of mol of H2o formed = 6.2 mole H2O

answer: 6.2 moles of H2O are formed; NaOH is the limiting reactant

2 H2 + O2 → 2 H2O

no of mol of O2 = 21/32 = 0.656 mol

no of mol of H2 = 6.5/2 = 3.25 mol

theoretical yield = 0.656*2 = 1.312 mol

actualyield = 17/18 = 0.944 mol

% yield = actualyield /theoretical yield*100

        = 0.944 / 1.312*100

       = 72%