Question #e03d5

1 Answer
Apr 23, 2017

Answer:

#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#