Question

5.20 g of H2 is ignited in a vessel containing 52.25 g O2 to produce H2O as the only product

a. Write a balanced equation

b. Calculate the number of moles of H2O produced by 5.20 g of H2

c. Calculate the number of moles produced by 52.25 g of O2

d. Identify the limiting reactant

e. Calculate the total mass of H2O produced in this reaction

Answer 1

The balanced equation is

2 H2 + O2 ------> 2 H2O

Number of moles of H2 = 5.20 g / 2.016 g/mol = 2.58 mole

number of moles of O2 = 52.25 g / 32.0 g/mol = 1.63 mole

b) from the balanced equation we can say that

2 mole of H2 produces 2 mole of H2O so

2.58 mole of H2 will produce 2.58 mole of H2O

mass of 1 mole of H2O = 18.016 g

so the mass of 2.58 mole of H2O = 46.5 g

Therefore, the mass of H2O produced from 5.20 g of H2 would be 46.5 g

c) from the balanced equation we can say that

1 mole of O2 produces 2 mole of H2O so

1.63 mole of O2 will produce

= 1.63 mole of O2 *(2 mole of H2O / 1 mole of O2)

= 3.26 mole of H2O

mass of 1 mole of H2O = 18.016 g

so the mass of 3.26 mole of H2O = 58.7 g

Therefore, the mass of H2O produced from 52.25 g of O2 would be 58.7 g

d)

from the balanced equation we can say that

2 mole of H2 requires 1 mole of O2 so

2.58 mole of H2 will require

= 2.58 mole of H2 *(1 mole of O2 / 2 mole of H2)

= 1.29 mole of O2

But we have 1.63 mole of O2 which is in excess so H2 is the limiting reactant

e) from the balanced equation we can say that

2 mole of H2 produces 2 mole of H2O so

2.58 mole of H2 will produce 2.58 mole of H2O

mass of 1 mole of H2O = 18.016 g

so the mass of 2.58 mole of H2O = 46.5 g

Therefore, the mass of H2O produced would be 46.5 g