Question

4A)     In the reaction, 2 H₂O₂ → 2 H₂O + O₂ (3 pts)

a.      How many grams of water are produced from 16.0 moles of hydrogen peroxide?

b.     How many grams of water are produced from 50.0 g of H₂O₂?

5A)    In the reaction, 3NaOH (aq) + H₃PO₄ (aq) → Na₃PO₄ (aq) + 3HOH (l)

a.      Which reactant is the limiting reactant if there are 10.0 mol of NaOH and 10.0 mol of H₃PO₄?

b.     Find the theoretical yield of water in grams?

6A)     In the reaction, CO (g) + 2 H₂ (g) →CH₃OH (l)

a. Which reactant is the limiting reactant and theoretical yield if 150.0 g of CO and 50.0 g of H₂ are mixed and allowed to react?

b. What is the percent yield if you actually made 155.0 grams

Answer 1

Q4.

a)

ratio is 2 mol of H2O2 = 2 mol of H2O

so, 1:1

then, 16 mol of H2O2 should produce 16 mol of H2O

now, 1 mol of H2O = 18 g

so 16 mol of water = 16*18 = 288 g of H2O

b)

first, change to mol

1 mol of H2O2 = 34 g/mol

mol of H2O2 = mass/MW = 50/34 = 1.4705 mol

ratio is 1:1 so 1.4705 mol of H2O2 =1.4705 mol of H2O

mass ofwater = mol*MW = 1.4705*18 = 26.469 g of H2O

Q5.

ratio is 3:1

so 10 mol of H3PO4 requires 10xx3 = 30 mol of NaOH, which we do NOT have, therefore NaOH will limit the reaction

c)

theoretical yield -- from total mass expected

10 mol of NaOH should produce 10 mol of H2O

so

mass of H2O = 10*18 = 180 g of H2O expected

Q6.

limiting reactant depends mainly on mol amount

mol of CO = mass/MW = 150/28 = 5.357

mol of H2 = mass/MW = 50/2 = 25 mol

clearly, there is excess of H2

limiting reactant is CO

b.

% yield = actual amount / theoretical * 100%

theoretical amount of methanol -- 1 mol of CO = 1 mol of CH3OH

5.357 mol of CO = 5.357 mol of methanol

mass of methanol expected = mol*MW = 5.357*32 = 171.424 g of methanol theoretically

% yield = (155)/171.424 *100 = 90.42 %