The reaction between potassium chlorate and red phosphorous takes place when you strike a match on a matchbox. If you were to react 52.9 g of potassium chlorate $(K C l O_{3})$ with excess red phosphorous, what mass of tetraphosphorous decaoxide $(P_{4} O_{10})$ could be produced?

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Solution

Potassium chlorate $(K C l O_{3})$ is reacted with excess red $(P_{4})$ to produce tetraphosphorous decaoxide $(P_{4} O_{10})$ and potassium chloride $(K C l)$ .
${10 K C l O_{3}}_{(s)} + {3 P_{4}}_{(s)} ⟶ {3 P_{4} O_{10}}_{(s)} + 10 K C l_{(s)}$
$52.9 g m$ of $K C l O_{3} = \frac{52.9}{122.55} = 0.431$ moles of $K C l O_{3}$
Now, $10$ moles of $K C l O_{3}$ produces $3$ moles of $P_{4} O_{10}$
$∴$ The mass of $P_{4} O_{10}$ produced= $\frac{3}{10} \times 0.431 \times 283.89 = 36.71 g m$

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