# Question 11371

Jun 22, 2017

$50$ ${\text{cm}}^{3}$ ${\text{NH}}_{3}$

#### Explanation:

Let's first write the chemical equation for this reaction:

${\text{N"_2(g) + 3"H"_2(g) rightleftharpoons 2"NH}}_{3} \left(g\right)$

For these purposes, I'll assume the reaction is driven to completion, rather than there being an equilibrium mixture present. I'll also assume the conditions are standard temperature and pressure.

Let's first find the number of moles of each reactant present, knowing that at STP, $1$ $\text{mol}$ $= 22.4$ $\text{L}$:

25cancel("cm"^3"N"_2)((1cancel("dm"^3))/(10^3cancel("cm"^3)))((1cancel("L"))/(1cancel("dm"^3)))((1"mol")/(22.4cancel("L"))) = 0.00112 ${\text{mol N}}_{2}$

75cancel("cm"^3"H"_2)((1cancel("dm"^3))/(10^3cancel("cm"^3)))((1cancel("L"))/(1cancel("dm"^3)))((1"mol")/(22.4cancel("L"))) = 0.00335 ${\text{mol H}}_{2}$

If you do the calculations, you'll see that (ideally) there is no limiting reactant here; both are assumptuously used up equally. We can therefore calculate the relative number of moles of ${\text{NH}}_{3}$ using either nitrogen or hydrogen:

$0.00112$ cancel("mol N"_2)((2color(white)(l)"mol NH"_3)/(1cancel("mol N"_2))) = 0.00223 ${\text{mol NH}}_{3}$

$0.00335$ cancel("mol H"_2)((2color(white)(l)"mol NH"_3)/(3cancel("mol H"_2))) = 0.00223 ${\text{mol NH}}_{3}$

The volume of ammonia produced (if the reaction goes to completion) is thus

$0.00223$ cancel("mol NH"_3)((22.4cancel("L"))/(1cancel("mol")))((1cancel("dm"^3))/(1cancel("L")))((10^3"cm"^3)/(1cancel("dm"^3)))

= color(red)(50 color(red)("cm"^3 color(red)("NH"_3#