Question #e03d5

1 Answer
Apr 23, 2017

56.9" g of NH"_3

Explanation:

Given:

(75" L of NH"_3)/1

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

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

Please observe who the Liters cancel and we are left with only moles of "NH"_3:

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

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

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

(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:

(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))

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