Question

An amount of 14.68g of hydrogen gas is allowed to react with 57.9g of oxygen gas to produce water. How many grams of water are produced? whch reactant is limiting? How much of the excess reactant remains?

Answer 1

Reaction between hydrogen and oxygen

H2 + 1/2 O2 - - - - > H2O

By stoichiometry,

1 mol of H2 reacts with 1/2 mol of O2 produces 1 mol of H2O.

2 g H2 reacts with [(32/2) = 16 g] of O2 produces 18 g H20.

By given amount,

Moles of hydrogen gas = 14.68 g / 2 g/mol = 7.34 mol

Moles of oxygen gas = 57.9 g / 32g/mol = 1.81 mol

7.34 moles of H2 reacts with (1.81/2 = 0.905) Moles of O2 produces 7.34 mol of H2O.

Grams of water = 7.34 moles * 18 g/mol = 132.12 g

H2 is limiting reagent and O2 is excess reagent.

Excess reactant (moles of O2) remains

= initially moles present - O2 used

= 1.81 - 0.905 = 0.905 Moles of O2