Question

Help with chemistry problem 6 please and how to put values in equation in the brackets underneath of the problem (so I can show my work)

Answer 1

Answer

449.4 grams

Explanation

The balanced chemical equation of the reaction is;

$N_2+3H_2\rightarrow2NH_3$

From the balanced chemical equation;

3 moles of H₂ reacted with 1 mole of N₂ to produce 2 moles of NH₃

Molar mass of H₂ = 2.016 g/mol

Molar mass of N₂ = 28.0134 g/mol

Molar mass of NH₃ = 17.031 g/mol

Convert mole to gram using the formula;

$Mole=\frac{\text{mass}}{\text{Molar mass}}$

For 1 mole N₂

$\begin{gathered} 1=\frac{\text{mass}}{28.0134} \\ mass=28.0134\text{ grams} \end{gathered}$

For 3 moles H₂

$\begin{gathered} 3=\frac{\text{mass}}{2.016} \\ m=3\times2.016=6.048\text{ grams} \end{gathered}$

For 2 moles NH₃

$\begin{gathered} 2=\frac{\text{mass}}{17.031} \\ m=2\times17.031=34.062\text{ grams} \end{gathered}$

We can now calculate, the mass of NH₃ that can be produced from 79.8 grams of H₂ as follows:

From the balanced equation we can say;

6.048 grams H₂ → 34.062 grams NH₃

∴ 79.8 grams H₂ → x grams NH₃

$\begin{gathered} \text{x grams }NH_3=\frac{34.062\times79.8}{6.048} \\ \text{x grams }NH_3=\frac{2718.1476}{6.048} \\ \text{x grams }NH_3=449.4291667\text{ grams} \\ \text{x grams }NH_3=449.4\text{ grams} \end{gathered}$

Therefore, 449.4 grams of Ammonia is produced if you started with 79.8 grams of Hydrogen.