Question

What volume of `O_2` at 912 mmHg and 21° C is required to synthesize 18.0 mol of `NO`?

Answer 1

`V_(O_2)=181L`

Explanation:

Assuming the nitrogen monoxide is made from oxygen and nitrogen according to the following reaction:

`N_2(g)+O_2(g)->2NO(g)`

Therefore, 2 moles of `NO` are prepared using only 1 mole of `O_2`, thus, we will need 9.00 moles of `O_2` to prepare 18.0 moles of `NO`.

The pressure of `O_2` is `P=912cancel(mmHg)xx(1atm)/(760cancel(mmHg)=1.20atm`

The Kelvin temperature is `T=21+273=294K`

Therefore, using the ideal gas law, `PV=nRT`, the volume of oxygen can be found by:

`V=(nRT)/P=(9.00cancel(mol)xx0.0821(cancel(atm)*L)/(cancel(mol)*cancel(K))xx294cancel(K))/(1.2cancel(atm))=181L`