chemical equation:

N$_{2}$+3H$_{2}$→2NH$_{3}$

 3,2    5,4  (mole)

set the ratio:

$\frac{n_{N2}}{1}$= $\frac{3,2}{1}$> $\frac{n_{O2}}{3}$= $\frac{5,4}{3}$

⇒After the reaction, N2 is left over, H2 is gone

⇒n$_{NH3}$=$\frac{2}{3}$.n$_{H2}$= $\frac{2}{3}$.5,4=3,6(mole) 

⇒n$_{N2 left}$=n$_{N2 entry}$-n$_{N2 react}$=3,2-5,4:3=1,4(mole)

Đáp án + Giải thích các bước giải:

 Balanced chemical equation:

$N_2 + 3H_2 \xrightarrow[Fe, Pt]{t^o} 2NH_3$$\\$`\text{1,8     5,4               3,6}`

Because $\dfrac{3,2}{1} > \dfrac{5,4}{3}$ so $H_2$ is the limiting reactant, and $N_2$ is the excess reactant. So we calculate the moles of reactants and the products by the number of $H_2$ moles, and from that we can find out that the greatest amount of $NH_3$ that can be made with given reactants are $3,6$ moles.

The leftover amount of $N_2$ is $3,2 - 1,8 = 1,4$ moles.