Consider the following chemical equation.
2H2 + O2 +2H20

a) mass of H₂O  produced = 4.0665 * 18 = 73.197 g

b)mass of H₂O  produced = 0.214 * 18 = 3.852 g

c) Oxygen gas, O₂, is the limiting reactant, since the number of moles available for reaction is far smaller and the mass of water produced from it is smaller too.

The reaction is the combustion of hydrogen gas to form water vapor. The question is not complete. A related question is given below:

Consider the following balanced equation:  

2H2 + O2 > 2H2O  

If you start with 8.133 g of H2 and 3.425 g of O2, find the following:

a) With excess O2, what mass (grams) of H2O would be produced by the H2?

b) With excess H2, what mass (grams) of H2O would be produced by the O2?

c) What is the chemical formula for the limiting reactant?

Explanation:

Equation of reaction: 2H₂ + O₂ ---> 2H₂O

From the equation of reaction, 2 moles of hydrogen gas reacts with 1 mole of oxygen gas to produce two moles of water

Molar mass of hydrogen gas, H₂ = 2 g/mol; molar mass of oxygen gas, O₂ = 32 g/mol; molar mass of H₂O = 18 g/mol

a) number of moles of hydrogen in 8.133 g = 8.133 g/ 2g/mol = 4.0665 moles

mole ratio of H₂ to H₂O is 1:1, therefore, 4.0665 moles of H₂O will be produced

mass of H₂O  produced = 4.0665 * 18 = 73.197 g

b) number of moles of oxygen in 3.425 g = 3.425 g /32 g/mol = 0.107 moles

mole ratio of O₂ to H₂O is 1:2, therefore, 0.107 * 2 moles of H₂O will be produced = 0.214 moles of H₂O

mass of H₂O  produced = 0.214 * 18 = 3.852 g

c) number of moles of hydrogen in 8.133 g = 4.0665 moles

number of moles of oxygen in 3.425 g = 0.107 moles

mole ratio of H₂ to O₂  = 4.0665/0.107 = 38 : 1

Oxygen gas, O₂, is the limiting reactant, since the number of moles available for reaction is far smaller and the mass of water produced from it is smaller too.