# Question e03d5

Apr 23, 2017

$56.9 {\text{ g of NH}}_{3}$

#### Explanation:

Given:

$\frac{75 {\text{ L of NH}}_{3}}{1}$

Assuming Standard Temperature and Pressure, we use the conversion factor for 1 mole is 22.4 Liters:

$\left(75 {\text{ L of NH"_3)/1(1" mole of NH"_3)/(22.4" L of NH}}_{3}\right)$

Please observe who the Liters cancel and we are left with only moles of ${\text{NH}}_{3}$:

$\left(75 \cancel{{\text{ L of NH"_3))/1(1" mole of NH"_3)/(22.4cancel(" L of NH}}_{3}}\right)$

Next we look up the molar mass for $\text{NH"_3}$ and write as a conversion factor from moles to grams:

$\left(75 \cancel{{\text{ L of NH"_3))/1(1" mole of NH"_3)/(22.4cancel(" L of NH}}_{3}}\right)$:

(75cancel(" L of NH"_3))/1(1" mole of NH"_3)/(22.4cancel(" L of NH"_3))(17.0" g of NH"_3)/(1" mole of NH"_3)

Again, please observe how the units cancel and we are left with only grams of ammonia gas:

$\left(75 \cancel{{\text{ L of NH"_3))/1(1cancel(" mole of NH"_3))/(22.4cancel(" L of NH"_3))(17.0" g of NH"_3)/(1cancel(" mole of NH}}_{3}}\right)$

We merely perform the multiplication and division:

(75cancel(" L of NH"_3))/1(1cancel(" mole of NH"_3))/(22.4cancel(" L of NH"_3))(17.0" g of NH"_3)/(1cancel(" mole of NH"_3)) = 56.9" g of NH"_3#