Question

Ammonia and oxygen react to form nitrogen and water. $$ 4 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ a. How many grams of \(\mathrm{O}_{2}\) are needed to react with \(13.6 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) ? b. How many grams of \(\mathrm{N}_{2}\) can be produced when \(6.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) reacts? c. How many grams of water are formed from the reaction of \(34.0 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) ?

Short answer

a. 19.2 g of \(\mathrm{O}_2\) b. 3.79 g of \(\mathrm{N}_2\) c. 54.0 g of water.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation is given by: \[ 4 \mathrm{NH}_{3}(g) + 3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{~N}_{2}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g) \]

Calculate the molar masses

Determine the molar masses of all involved substances:\[\text{Molar mass of } \mathrm{NH}_{3} = (1 \times 14) + (3 \times 1) = 17 \text{ g/mol} \]\[\text{Molar mass of } \mathrm{O}_{2} = (2 \times 16) = 32 \text{ g/mol} \]\[\text{Molar mass of } \mathrm{N}_{2} = (2 \times 14) = 28 \text{ g/mol} \]\[\text{Molar mass of } \mathrm{H}_{2} \mathrm{O} = (2 \times 1) + (1 \times 16) = 18 \text{ g/mol} \]

Part A - Step 1: Calculate moles of \( \mathrm{NH}_{3} \)

Using the mass given for \(13.6\text{ g of }\mathrm{NH}_{3}\):\[ \text{Moles of } \mathrm{NH}_{3} = \frac{13.6 \text{ g}}{17 \text{ g/mol}} = 0.8 \text{ mol} \]

Part A - Step 2: Determine moles of \( \mathrm{O}_2 \) needed

Using the balanced equation, the mole ratio of \(\mathrm{NH}_{3}\) to \(\mathrm{O}_2\) is 4:3:\[ \text{Moles of } \mathrm{O}_2 \text{ needed} = 0.8 \text{ mol } \mathrm{NH}_{3} \times \frac{3 \text{ mol } \mathrm{O}_2}{4 \text{ mol } \mathrm{NH}_{3}} \ = 0.6 \text{ mol } \mathrm{O}_2 \]

Part A - Step 3: Calculate the mass of \( \mathrm{O}_2 \) needed

Using the molar mass of \(\mathrm{O}_2\), find the mass:\[ \text{Mass of } \mathrm{O}_2 = 0.6 \text{ mol} \times 32 \text{ g/mol} = 19.2 \text{ g} \]

Part B - Step 1: Calculate moles of \( \mathrm{O}_2 \)

Using the mass given for \(6.50\text{ g of } \mathrm{O}_2\):\[ \text{Moles of } \mathrm{O}_2 = \frac{6.50 \text{ g}}{32 \text{ g/mol}} = 0.203125 \text{ mol} \]

Part B - Step 2: Determine moles of \( \mathrm{N}_2 \) produced

Using the balanced equation, the mole ratio of \(\mathrm{O}_2\) to \(\mathrm{N}_2\) is 3:2:\[ \text{Moles of } \mathrm{N}_2 \text{ produced} = 0.203125 \text{ mol } \mathrm{O}_2 \times \frac{2 \text{ mol } \mathrm{N}_2}{3 \text{ mol } \mathrm{O}_2} = 0.1354167 \text{ mol } \mathrm{N}_2 \]

Part B - Step 3: Calculate the mass of \( \mathrm{N}_2 \) produced

Using the molar mass of \(\mathrm{N}_2\):\[ \text{Mass of } \mathrm{N}_2 = 0.1354167 \text{ mol} \times 28 \text{ g/mol} = 3.7916667 \text{ g} \]

Part C - Step 1: Calculate moles of \( \mathrm{NH}_3 \)

Using the mass given for \(34.0\text{ g of } \mathrm{NH}_3\):\[ \text{Moles of } \mathrm{NH}_3 = \frac{34.0 \text{ g}}{17 \text{ g/mol}} = 2.0 \text{ mol} \]

Part C - Step 2: Determine moles of water produced

Using the balanced equation, the mole ratio of \(\mathrm{NH}_3\) to \(\mathrm{H}_2 \mathrm{O}\) is 4:6:\[ \text{Moles of water produced} = 2.0 \text{ mol } \mathrm{NH}_3 \times \frac{6 \text{ mol } \mathrm{H}_2 \mathrm{O}}{4 \text{ mol } \mathrm{NH}_3} = 3.0 \text{ mol } \mathrm{H}_2 \mathrm{O} \]

Part C - Step 3: Calculate the mass of water produced

Using the molar mass of \(\mathrm{H}_2\mathrm{O}\):\[ \text{Mass of water} = 3.0 \text{ mol} \times 18 \text{ g/mol} = 54.0 \text{ g} \]

Key concepts

Balanced Chemical Equations

To solve any stoichiometry problem, we start with a balanced chemical equation. This equation shows the reactants and products in the reaction, and the proportion in which they react. Balancing involves ensuring the number of atoms of each element is the same on both sides of the equation. For instance, in our example reaction: \[4 \mathrm{NH}_{3}(g) + 3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{~N}_{2}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g) \] we see that 4 molecules of ammonia (\