Question #e81fc

Apr 25, 2016

${\text{0.840 moles NH}}_{3}$

Explanation:

The key to this problem is the balanced chemical equation that describes this synthesis reaction

${\text{N"_ (2(g)) + 3"H"_ (2(g)) -> color(red)(2)"NH}}_{3 \left(g\right)}$

The balanced chemical equation tells you that when $1$ molecule of nitrogen gas, ${\text{N}}_{2}$, reacts with $3$ molecules of hydrogen gas, ${\text{H}}_{2}$, the reaction produces $\textcolor{red}{2}$ molecules of ammonia, ${\text{NH}}_{3}$.

Now, a mole is simply a very, very large collection of molecules. More specifically, one mole is defined as

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{1 mole" = 6.022 * 10^(23)"molecules} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ Avogadro's number

This means that every time you have $6.022 \cdot {10}^{23}$ molecules of a substance, you have one mole of that substance. This means that the molecule ratio that is described by the balanced chemical equation will be equivalent to a mole ratio.

So, the reaction will consume $1$ mole of nitrogen gas for every $3$ moles of hydrogen gas and produce $\textcolor{red}{2}$ moles of ammonia.

Since the problem doesn't mention the number of moles of nitrogen gas available for the reaction, you can assume that this reactants is in excess, i.e. you have more than you need to ensure that all the moles of hydrogen gas react.

Use the $3 : \textcolor{red}{2}$ mole ratio that exists between hydrogen and ammonia to determine how many moles of the latter will be formed by the reaction

$1.26 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles H"_2))) * (color(red)(2)color(white)(a)"moles NH"_3)/(3color(red)(cancel(color(black)("moles H"_2)))) = color(green)(|bar(ul(color(white)(a/a)"0.840 moles NH}}_{3} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the number of moles of hydrogen gas.