Question

If 5.00 g of hydrogen gas and 5.00 g of oxygen gas react at STP according to the balanced equation2H

2H2(g)+O2 (g)→2H2 O(g)

a) Determine the limiting reactant:

The limiting reactant is ______ (Hydrogen gas or Oxygen gas? Explain your choice.)
b) Identify the excess reactant:

The excess reactant is ______ (Hydrogen gas or Oxygen gas? Explain your choice.)
c) Calculate the theoretical yield of water vapor:

Answer 1

The limiting reactant is oxygen gas. The excess reactant is hydrogen gas. The theoretical yield of water vapor is 0.3125 mol.

Step-by-step explanation:

The balanced equation for the reaction is:
2H2(g) + O2(g) → 2H2O(g)
To determine the limiting reactant, we need to calculate the number of moles of hydrogen gas and oxygen gas, and then compare them to the stoichiometric ratio in the balanced equation. The molar mass of hydrogen gas is 2 g/mol and the molar mass of oxygen gas is 32 g/mol.
For hydrogen gas: 5 g / 2 g/mol = 2.5 mol
For oxygen gas: 5 g / 32 g/mol ≈ 0.15625 mol
Since the stoichiometric ratio is 2:1 for hydrogen gas to oxygen gas, and we have more moles of hydrogen gas than oxygen gas, the limiting reactant is oxygen gas.

To identify the excess reactant, we can subtract the amount of the limiting reactant used from the amount of reactant initially provided.
Using the mole ratio from the balanced equation, we can calculate the amount of water produced from the limiting reactant, which is oxygen gas. 0.15625 mol of O2 will produce 2*0.15625 mol of H2O, which is 0.3125 mol.
Therefore, the theoretical yield of water vapor is 0.3125 mol.