Answer to Question #95573 in General Chemistry for Tea

Question

The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. If you were to react 48.7 g of potassium chlorate (KClO3) with excess red phosphorus, what mass of tetraphosphorus decaoxide (P4O10) could be produced?

Answer 1

The equation of chemical reaction (unbalanced!) is

$$KCl{O_3} + P \to {P_4}{O_{10}} + KCl$$

It's easy to balance it (first balance the $O$ multiplying $KCl{O_3}$ by 10 and ${P_4}{O_{10}}$ by 3, then balance $K$ multiplying $KCl$ by 10 and balance $P$ multiplying $P$ by 12, $Cl$ will be balanced automatically) and we get

$$10KCl{O_3} + 12P \to 3{P_4}{O_{10}} + 10KCl$$

Then calculate the molar mass of $KCl{O_3}$ as the sum of atomic weights (can be found in periodic table) ${M_{KCl{O_3}}} = {M_K} + {M_{Cl}} + 3{M_O}$ , we get

$${M_{KCl{O_3}}} \approx 39.098[\frac{{\text{g}}}{{{\text{mol}}}}] + 35.446[\frac{{\text{g}}}{{{\text{mol}}}}] + 3 \cdot 16[\frac{{\text{g}}}{{{\text{mol}}}}] = 122.544[\frac{{\text{g}}}{{{\text{mol}}}}]$$

The same for ${P_4}{O_{10}}$

$${M_{{P_4}{O_{10}}}} \approx 4 \cdot 30.974[\frac{{\text{g}}}{{{\text{mol}}}}] + 10 \cdot 16[\frac{{\text{g}}}{{{\text{mol}}}}] = 283.896[\frac{{\text{g}}}{{{\text{mol}}}}]$$

Now, we can see from the equation that for each 10 moles of $KCl{O_3}$ we get 3 moles of ${P_4}{O_{10}}$ , in other words for each 1 mole of $KCl{O_3}$ we get $\frac{3}{{10}}$ moles of ${P_4}{O_{10}}$ . We can calculate the number of moles of $KCl{O_3}$

$${\nu _{KCl{O_3}}} = \frac{{48.7[{\text{g}}]}}{{122.544[\frac{{\text{g}}}{{{\text{mol}}}}]}} \approx 0.397[{\text{mol}}]$$

And we can calculate the number of moles obtained ${P_4}{O_{10}}$

$${\nu _{{P_4}{O_{10}}}} = \frac{3}{{10}}{\nu _{KCl{O_3}}} = \frac{3}{{10}} \cdot 0.397[{\text{mol}}] \approx 0.12[{\text{mol}}]$$

Let's calculate the mass of obtained ${P_4}{O_{10}}$

$${m_{{P_4}{O_{10}}}} = {\nu _{{P_4}{O_{10}}}} \cdot {M_{{P_4}{O_{10}}}} = 0.12[{\text{mol}}] \cdot 283.896[\frac{{\text{g}}}{{{\text{mol}}}}] \approx 34.07[{\text{g}}]$$