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Exercise · Q22

Q.4.0 g4.0\ \text{g} of hydrogen gas is mixed with 40.0 g40.0\ \text{g} of oxygen gas and ignited to form water, 2H2(g)+O2(g)→2H2O(g)2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(g). Identify the limiting reagent and calculate the mass of water formed.

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Balanced equation: 2H2(g)+O2(g)→2H2O(g)2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(g).

Moles of H2=4.02.016=1.984 mol\text{H}_2 = \dfrac{4.0}{2.016} = 1.984\ \text{mol}

Moles of O2=40.032.00=1.25 mol\text{O}_2 = \dfrac{40.0}{32.00} = 1.25\ \text{mol}

Divide by coefficients: H2\text{H}_2: 1.9842=0.992\dfrac{1.984}{2} = 0.992; O2\text{O}_2: 1.251=1.25\dfrac{1.25}{1} = 1.25.

Since 0.992<1.250.992 < 1.25, hydrogen gives the smaller value and is the limiting reagent — it will be completely used up first.

Mass of water formed (using the limiting reagent, H2\text{H}_2, with the 1:11:1 mole ratio between H2\text{H}_2 and H2O\text{H}_2\text{O}):

n(H2O)=n(H2)=1.984 moln(\text{H}_2\text{O}) = n(\text{H}_2) = 1.984\ \text{mol} …

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