Question

Ammonia is produced by the chemical reaction of hydrogen and nitrogen.

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

a. How many moles of H₂ are needed to react with 1.8 moles of N₂?

b. How many moles of N₂ reacted if 0.60 moles of NH₃ is produced?

c. How many moles of NH₃ are produced when 1.4 moles of H₂ react?

Short answer

a. 5.4 moles of H₂ are needed to react with 1.8 moles of N₂ .

b.0.3 moles of N₂ reacted if 0.60 moles of NH₃ is produced.

c.0.933 moles of NH₃ are produced when 1.4 moles of H₂ react.

Step-by-step solution

Given information

The moles of N₂ are 1.8 moles.

Part (a) Explanation

The balanced chemical reaction is given as follows:

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

From the equation, it is observed that 1 mole of nitrogen gas reacts with three moles of hydrogen to produce 2 moles of ammonia.

Thus, to calculate the number of moles of hydrogen needed if 1.8 moles of Nitrogen react, the following conversation takes place.

$\frac{3\mathrm{moles}{\mathrm{H}}_2}{1\mathrm{mole}{\mathrm{N}}_2}\mathrm{and}\frac{1\mathrm{mole}{\mathrm{N}}_2}{3\mathrm{moles}{\mathrm{H}}_2}$

Therefore, the number of moles of hydrogen are:

$1.8\cancel{\mathrm{moles}{\mathrm{N}}_2}\times \frac{3\mathrm{mole}s{\mathrm{H}}_2}{1\cancel{\mathrm{mole}{\mathrm{N}}_2}}=1.8\times 3=5.4\mathrm{mole}s{\mathrm{H}}_2$

Hence, 5.4 moles of H₂ are needed to react with 1.8 moles of N₂ .

Part (b) Given Information

The moles of ammonia that are produced are 0.60 moles.

Part (b) Explanation

The balanced chemical reaction is given as follows:

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

From the equation, it is observed that 1 mole of nitrogen gas reacts with three moles of hydrogen to produce 2 moles of ammonia.

Thus, to calculate the number of moles of nitrogen needed if 0.60 moles of ammonia is produced, the following conversation takes place.

$\frac{1\mathrm{mole}{\mathrm{N}}_2}{2\mathrm{moles}{\mathrm{NH}}_3}\mathrm{and}\frac{2\mathrm{moles}{\mathrm{NH}}_3}{1\mathrm{mole}{\mathrm{N}}_2}$

Therefore, if 0.60 moles of NH₃ is produced, the moles of nitrogen will be:

$0.60\cancel{\mathrm{moles}{\mathrm{NH}}_3}\times \frac{1\mathrm{mole}{\mathrm{N}}_2}{2\cancel{\mathrm{moles}{\mathrm{NH}}_3}}=0.3\mathrm{moles}{\mathrm{NH}}_3$

Hence, 0.3 moles of N₂ reacted if 0.60 moles of NH₃ is produced.

Part (c) Given Information

The moles ofH₂ are 1.4 moles.

Part (c) Explanation

The balanced chemical reaction is given as follows:

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

From the equation, it is observed that 1 mole of nitrogen gas reacts with three moles of hydrogen to produce 2 moles of ammonia.

Thus, to calculate the number of moles of ammonia needed if 1.4 moles of hydrogen react, the following conversation takes place.

$\frac{3\mathrm{moles}{\mathrm{H}}_2}{2\mathrm{moles}{\mathrm{NH}}_3}\mathrm{and}\frac{2\mathrm{moles}{\mathrm{NH}}_3}{3\mathrm{moles}{\mathrm{H}}_2}$

Therefore, for 1.4 moles ofH₂ , the moles of ammonia are:

$1.4\cancel{\mathrm{moles}{\mathrm{H}}_2}\times \frac{2\mathrm{moles}{\mathrm{NH}}_3}{3\cancel{\mathrm{moles}{\mathrm{H}}_2}}=\frac{1.4\times 2\mathrm{moles}{\mathrm{NH}}_3}{3}=0.933\mathrm{moles}{\mathrm{NH}}_3$

Hence, 0.933 moles of NH₃ are produced when 1.4 moles of H₂ react.