The mass of P4O10 that will be obtained from the reaction of 1.33 gram of P4 and 5.07 of oxygen is
Answer: B
The answer is 3.05 g.
The reaction is \[\mathrm{P_4 + 5\,O_2 \rightarrow P_4O_{10}}\]
Moles of each reactant: \[n(\mathrm{P_4})=\frac{1.33}{124}=0.01073\ \text{mol},\qquad n(\mathrm{O_2})=\frac{5.07}{32}=0.1584\ \text{mol}\]
\(0.01073\) mol of \(\mathrm{P_4}\) needs \(5\times0.01073=0.0536\) mol of \(\mathrm{O_2}\), and far more is available, so \(\mathrm{P_4}\) is the limiting reagent.
One mole of \(\mathrm{P_4}\) gives one mole of \(\mathrm{P_4O_{10}}\) \((M=284\ \mathrm{g\,mol^{-1}})\): \[m=0.01073\times284=3.05\ \mathrm{g}\]
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