Question

if 2.05 mol of h₂ and 1.55 mol of o₂ react, how many moles of h₂o can be produced in the reaction below? 2h₂(g) + o₂(g) → 2h₂o(g)

Explanation:

Step1: Determine the limiting reactant

From the balanced equation \(2H_{2}(g)+O_{2}(g)\to2H_{2}O(g)\), the mole ratio of \(H_{2}\) to \(O_{2}\) is \(2:1\).
If we have \(n(H_{2}) = 2.05\space mol\), the amount of \(O_{2}\) required for complete reaction of \(H_{2}\) is \(n_{req}(O_{2})=\frac{2.05\space mol}{2}=1.025\space mol\).
We have \(n(O_{2}) = 1.55\space mol\), which is more than \(1.025\space mol\). So \(H_{2}\) is the limiting reactant.

Step2: Calculate the moles of \(H_{2}O\) produced

Using the mole ratio of \(H_{2}\) to \(H_{2}O\) (which is \(2:2 = 1:1\) from the balanced equation).
Since \(H_{2}\) is limiting, \(n(H_{2}O)=n(H_{2})\) (based on the stoichiometry of the reaction).

Answer:

\(2.05\space mol\)