Recognizing that 1 mole of an ideal gas occupies 22.41 L at STP and using the molar mass of NH_{3} (17.03 g), we write the sequence of conversions as

grams of  NH_{3} → moles of  NH_{3} → liters of  NH_{3} at STP

so the volume of  NH_{3} is given by

V=7.40  \cancel{g  NH_{3}}\times \frac{1  \cancel{mol  NH_{3}}}{17.03  \cancel{g  NH_{3}}} \times \frac{22.41  L}{1  \cancel{mol  NH_{3}}}

=9.74  L

It is often true in chemistry, particularly in gas law calculations, that a problem can be solved in more than one way. Here the problem can also be solved by first converting 7.40 g of NH_{3} to number of moles of   NH_{3}, and then applying the ideal gas equation (V = nRT⁄P). Try it.

Check Because 7.40 g of  NH_{3} is smaller than its molar mass, its volume at STP should be smaller than 22.41 L. Therefore, the answer is reasonable.