# #60*g# of ammonia gas, and #2.5*mol# dihydrogen gas, are confined in a piston at #"STP"#. What is the volume of the piston?

##### 2 Answers

Treat both gases as ideal and use the ideal gas equation. Note that idealization of gases is not always justified, but here, with moderate pressures (1 atm) and moderate temperatures, it is entirely reasonable.

#### Explanation:

Volume of hydrogen: **57 L**

Volume of ammonia: **80 L**

#### Explanation:

Notice that your gases are at **STP** conditions, which means that you can use the molar volume of a gas at STP to make your life easier.

**STP** conditions imply a pressure of **one mole** of any idea lgas aoccupies exactly **22.7 L**.

So basically all you need to do is figure out how many moles of each gas you have. You already know how many moles of hydrogen gas you have, so you can calculate its volume to be

#2.5color(red)(cancel(color(black)("moles"))) * "22.7 L"/(1color(red)(cancel(color(black)("mole")))) = "56.75 L"#

To get the number of moles of ammonia, use its molar mass

#60color(red)(cancel(color(black)("g"))) * ("1 mole NH"""_3)/(17.03color(red)(cancel(color(black)("g")))) = "3.52 moles NH"""_3#

This means that the volume of the ammonia sample will be

#3.52color(red)(cancel(color(black)("moles"))) * "22.7 L"/(1color(red)(cancel(color(black)("mole")))) = "79.9 L"#

You need to round the answers to two sig figs for hydrogen and to one sig fig for ammonia

#V_(H_2) = color(green)("57 L")" "# and#" "V_(NH_3) = color(green)("80 L")#

**SIDE NOTE** *Many textbooks and online sources still list the molar volume of a gas at STP as being equal to 22.4 L.*

*That value isbased on the old definition of STP, which implied a pressure of* *and a temperature of*

*If this was the value you were supposed to use, simply redo the calculations using 22.4 instead of 22.7 L.*