Question

5K + KNO3 ------ 3K2O + N

Answer 1

Answer:

174,957.143 grams of potassium and 89,228.478 grams of potassium nitrate will be needed.

Explanation:

$5K +KNO_3\rightarrow 3K_2O + N$

Mass of nitrogen =  27 lbs = 12,247 g

1 lbs = 453.592 g

Moles of nitrogen = $\frac{12,247 g}{14 g/mol}=874.786 mol$

According to reaction, 1 mole of nitrogen is produced from 5 moles of potassium and 1 mole of potassium nitrate.

Then 874.786 mol of nitrogen will be obtained from :

$\frac{5}{1}\times 874.786 mol=4,373.928 mol$ of potassium.

Then 874.786 mol of nitrogen will be obtained from :

$\frac{1}{1}\times 874.786 mol=874.789 mol$ of potassium nitrate.

Mass of 4,373.928 moles of potassium:

$4,373.928 mol\times 40 g/mol=174,957.143 g$ of potassium

Mass of 874.789 moles of potassium nitrate:

$874.789 mol\times 102 g/mol=89,228.478 g$ of potassium nitrate