# Question 3c82d

Aug 4, 2015

Mass of calcium nitrate: 1200 kg

#### Explanation:

The idea behind this problem is that you first need to determine how much nitrogen is supplied per mole of calcium nitrate, then use this value to determine how much you'd get per gram of calcium nitrate.

In other words, you need to determine the percent composition of nitrogen in $C a {\left(N {O}_{3}\right)}_{2}$. To do that, use the molar mass of nitrogen and the molar mass of calcium nitrate.

This important thing to notice is that you get 2 moles of nitrogen per mole of calcium nitrate, which means that you have

(2 * 14.007color(red)(cancel(color(black)("g/mol"))))/(164.09color(red)(cancel(color(black)("g/mol")))) * 100 = 17.1%

This means that, for every 100g of calcium nitrate, you get 17.1 g of nitrogen.

Since you're dealing with a 10-hectare field, you need to supply a total of

10color(red)(cancel(color(black)("hectares"))) * "20 kg nitrogen"/(1color(red)(cancel(color(black)("hectare")))) = "200 kg nitrogen"

You can express this value in grams to get

200color(red)(cancel(color(black)("kg"))) * "1000 g"/(1color(red)(cancel(color(black)("kg")))) = 200 * 10^3"g"

Since you need 100g of calcium nitrate to get 17.1 g of nitrogen, you can determine that

200 * 10^3color(red)(cancel(color(black)("g nitrogen"))) * ("100 g "Ca(NO_3)_2)/(17.1color(red)(cancel(color(black)("g nitrogen")))) = 1169.6 * 10^3"g" $C a {\left(N {O}_{3}\right)}_{2}$

You can convert this back to kilograms if you want

1169.6 * 10^3color(red)(cancel(color(black)("g"))) * (10^(-3)"kg")/(1color(red)(cancel(color(black)("g")))) = "1169.6 kg"# $C a {\left(N {O}_{3}\right)}_{2}$

Now, I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the mass of nitrogen and area of the field.

${m}_{C a {\left(N {O}_{3}\right)}_{2}} = \textcolor{g r e e n}{\text{1200 kg}}$