Question

The reaction between potassium chlorate and red phosphorus is highly exothermic and takes place when you strike a match on a matchbox. If you were to react 52.9 g of potassium chlorate (KCl03) with red phosphorus, how many grams of tetraphosphorus decaoxide (P4010) would be produced?

Accepted Answer

1. The reaction is:  10  mathrm{KClO_3(s)} + 3  mathrm{P_4(s)}  rightarrow 3  mathrm{P_4O_{10}(s)} + 10  mathrm{KCl(s)}  2. Convert to moles:  n_{ mathrm{KClO_3}} =  frac{m_{ mathrm{KClO_3}}}{M_{ mathrm{KClO_3}}} =  frac{52.9}{122.55} = 0.432  text{ (mole)}  3. Multiply by mole ratio:  n_{P_4O_{10}} = n_{ mathrm{KClO_3}}  cdot  frac{3}{10} = 0.432  cdot 0.3 = 0.129  text{ (mole)}  4. Convert to grams:  m_{P_4O_{10}} = n_{P_4O_{10}}  cdot M_{P_4O_{10}} = 0.129  cdot 283.89 = 36.62  text{ (g)}  Answer: 36.62 g.

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