Question

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

How much in grams would you need of each reactant to make 27 lbs of nitrogen?

Answer 1

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

Step-by-step explanation:

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

Mass of nitrogen = 27 lbs = 12,247 g

1 lbs = 453.592 g

Moles of nitrogen =
`(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 :

`(5)/(1)* 874.786 mol=4,373.928 mol` of potassium.

Then 874.786 mol of nitrogen will be obtained from :

`(1)/(1)* 874.786 mol=874.789 mol` of potassium nitrate.

Mass of 4,373.928 moles of potassium:

`4,373.928 mol* 40 g/mol=174,957.143 g` of potassium

Mass of 874.789 moles of potassium nitrate:

`874.789 mol* 102 g/mol=89,228.478 g` of potassium nitrate