Question

it turns out that there are:
2 h₂(g) + o₂(g) → 2 h₂o(g)
10 grams (5 moles) of hydrogen
96 grams (3 moles) of oxygen
10 g h₂ + 96 g o₂ → h₂o(g)
5 moles h₂ 3 moles o₂
knowing what you do about the
moles of each of the reactants,
which of the following is the
limiting reactant?
h₂
o₂
h₂o
none of the above

Explanation:

Step1: Analyze the reaction ratio

From the balanced equation \(2H_{2}(g)+O_{2}(g)\to2H_{2}O(g)\), the mole ratio of \(H_{2}\) to \(O_{2}\) is \(2:1\).

Step2: Calculate the required moles of \(O_{2}\) for given \(H_{2}\)

If there are \(n = 5\) moles of \(H_{2}\), according to the ratio, the moles of \(O_{2}\) required is \(n_{O_{2}\text{(required)}}=\frac{5}{2}=2.5\) moles.

Step3: Compare with available \(O_{2}\)

We have \(n_{O_{2}\text{(available)}} = 3\) moles. Since \(2.5<3\), \(H_{2}\) will be completely consumed first.

Answer:

\(H_{2}\)