Answer
36.8 g of $P_4O_{10}$
Work Step by Step
1. Balance the equation:
- Balance the amount of oxygen in each side:
$$10 \space KClO_3(s)+P_4(s)\longrightarrow 3 \space P_4O_{10}(s) + KCl(s)$$
- Balance the amount of potassium:
$$10 \space KClO_3(s)+P_4(s)\longrightarrow 3 \space P_4O_{10}(s) + 10 \space KCl(s)$$
- Balance the amount of phosphorus in each side:
$$10 \space KClO_3(s)+3 \space P_4(s)\longrightarrow 3 \space P_4O_{10}(s) + 10 \space KCl(s)$$
2. Find the mass of $P_4O_{10}$ produced:
$ KClO_3 $ : ( 35.45 $\times$ 1 )+ ( 16.00 $\times$ 3 )+ ( 39.10 $\times$ 1 )= 122.55 g/mol
$$ \frac{1 \space mole \space KClO_3 }{ 122.55 \space g \space KClO_3 } \space and \space \frac{ 122.55 \space g \space KClO_3 }{1 \space mole \space KClO_3 }$$
$ P_4O_{10} $ : ( 16.00 $\times$ 10 )+ ( 30.97 $\times$ 4 )= 283.88 g/mol
$$ \frac{1 \space mole \space P_4O_{10} }{ 283.88 \space g \space P_4O_{10} } \space and \space \frac{ 283.88 \space g \space P_4O_{10} }{1 \space mole \space P_4O_{10} }$$
$$ 52.9 \space g \space KClO_3 \times \frac{1 \space mole \space KClO_3 }{ 122.55 \space g \space KClO_3 } \times \frac{ 3 \space moles \space P_4O_{10} }{ 10 \space moles \space KClO_3 } \times \frac{ 283.88 \space g \space P_4O_{10} }{1 \space mole \space P_4O_{10} } = 36.8 \space g \space P_4O_{10} $$
