A mixture of 0.0375 g of hydrogen and 0.0185 mol of oxygen

Step 1 of 4

The balanced equation is

\(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)\)

Here, we are calculating the moles of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) from the moles of \(\mathrm{H}_{2}\) if we consider \(\mathrm{O}_{2}\) is limiting reagent:

Given that,

The mass of \(\mathrm{H}_{2}=0.0375 \mathrm{~g}\)

We know that,

The moles of \(\mathrm{H}_{2}=\left(0.0375 \mathrm{~g} \mathrm{H}_{2}\right) \times \frac{1.0 \mathrm{~mol} \mathrm{H}_{2}}{2.016 \mathrm{~g} \mathrm{H}_{2}}=0.01860 \mathrm{~mol} \mathrm{H}_{2}\)

Moles of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) from \(\mathrm{H}_{2}=\left(0.01860 \mathrm{~mol} \mathrm{H}_{2}\right) \times \frac{2.0 \text { molof } \mathrm{H}_{2} \mathrm{O}(l)}{2.0 \mathrm{~mol} \mathrm{H}_{2}}=0.01860 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\)

Thus, 0.01860 mol \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) is formed from 0.0375 g of hydrogen if we consider \(O_2\) is limiting reagent.