# How many molecules of NH_3 are produced from  7.52\times10^-4 g of  H_2?

Mar 17, 2018

I get $1.51 \cdot {10}^{20}$ molecules of ammonia.

#### Explanation:

We have the familiar Haber process equation...

${N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right) \stackrel{{450}^{\circ} \text{C")stackrel(200 \ "atm")stackrel(Fe \ "catalyst}}{r} i g h t \le f t h a r p \infty n s 2 N {H}_{3} \left(g\right)$

Since the mole ratio of ${H}_{2}$ to $N {H}_{3}$ is $3 : 2$, then three moles of hydrogen gas are required to produce two moles of ammonia gas.

We got: $7.52 \cdot {10}^{-} 4 \setminus \text{g of} \setminus {H}_{2}$, so let's convert that amount into moles.

${H}_{2}$ has a molar mass of $2 \setminus \text{g/mol}$. So here, we got

(7.52*10^-4color(red)cancelcolor(black)"g")/(2color(red)cancelcolor(black)"g""/mol")=3.76*10^-4 \ "mol"

Since we have $3.76 \cdot {10}^{-} 4$ mol of hydrogen, then we are gonna produce:

3.76*10^-4color(red)cancelcolor(black)("mol" \ H_2)*(2 \ "mol" \ NH_3)/(3color(red)cancelcolor(black)("mol" \ H_2))~~2.51*10^-4 \ "mol" \ NH_3

We know that one mole of molecules is equal to $6.02 \cdot {10}^{23}$ molecules. So here, we produce

$2.51 \cdot {10}^{-} 4 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol"*(6.02*10^23 \ "molecules")/(color(red)cancelcolor(black)"mol")~~1.51*10^20 \ "molecules}}}}$