Question

How many grams of H2 are needed to produce 10.71g of NH3?

Answer 1

Answer on Question#39097 - Chemisry - Inorganic Chemistry

Question

How many grams of $\mathrm{H}_2$ are needed to produce 10.71 g of $\mathrm{NH}_3$?

Solution:

$\mathrm{NH}_3$ can be produced by the reaction:

$$3 \mathrm{H}_2(\mathrm{g}) + \mathrm{N}_2(\mathrm{g}) = 2 \mathrm{NH}_3(\mathrm{g})$$

Molar mass of $\mathrm{H}_2$ equals:

$$M(\mathrm{H}_2) = 2 M(\mathrm{H}) = 2 \cdot 1 = 2 \frac{\mathrm{g}}{\text{mole}}$$

Mass of 3 moles of hydrogen equals:

$$3 \cdot 2 = 6 \mathrm{g}$$

Molar mass of $\mathrm{NH}_3$ equals:

$$M(\mathrm{NH}_3) = M(\mathrm{N}) + 3 M(\mathrm{H}) = 14 + 3 \cdot 1 = 17 \frac{\mathrm{g}}{\text{mole}}$$

Mass of 2 moles of $\mathrm{NH}_3$ equals:

$$2 \cdot 17 = 34 \mathrm{g}$$

Then we make a proportion:

- 6 g of $\mathrm{H}_2$ produce 34 g of $\mathrm{NH}_3$

- $x \text{ g of } \mathrm{H}_2 - 10.71 \text{ g of } \mathrm{NH}_3$

- $x = \frac{6 \cdot 10.71}{34} = 1.89 \mathrm{g}$

Answer: $m(\mathrm{H}_2) = 1.89 \mathrm{g}$