Question

Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) emits a large quantity of energy when it reacts with oxygen, which has led to hydrazine's use as a fuel for rockets: $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ How many moles of each of the gaseous products are produced when \(20.0 \mathrm{g}\) of pure hydrazine is ignited in the presence of \(20.0 \mathrm{g}\) of pure oxygen? How many grams of each product are produced?

Short answer

In conclusion, when 20.0 g of pure hydrazine is ignited in the presence of 20.0 g of pure oxygen, 0.624 moles (17.49 g) of nitrogen and 1.248 moles (22.50 g) of water are produced as gaseous products.

Step-by-step solution

1. Balance the chemical equation

We are given the following chemical equation: \[ \text{N}_2\text{H}_4(l) + \text{O}_2(g) \rightarrow \text{N}_2(g) + 2 \text{H}_2\text{O}(g) \] This equation is already balanced: 1 mole of hydrazine reacts with 1 mole of oxygen to produce 1 mole of nitrogen and 2 moles of water.

2. Convert mass of reactants to moles

We will use the molar mass of hydrazine and oxygen to convert their given masses to moles. Molar mass of N2H4 = (2 × 14.01 g/mol (N) + 4 × 1.01 g/mol (H)) = 32.05 g/mol Moles of N2H4 = (20.0 g) / (32.05 g/mol) = 0.624 moles Molar mass of O2 = (2 × 16.00 g/mol) = 32.00 g/mol Moles of O2 = (20.0 g) / (32.00 g/mol) = 0.625 moles

3. Determine the limiting reactant

As per the balanced equation, 1 mole of N2H4 reacts with 1 mole of O2. We have 0.624 moles of N2H4 and 0.625 moles of O2. Since we have slightly less N2H4, it is the limiting reactant, and thus determines the amounts of products formed.

4. Calculate moles of products formed

Using the stoichiometry of the balanced equation, we can find the moles of products formed. Moles of N2 = moles of N2H4 (limiting reactant) = 0.624 moles Moles of H2O = 2 × moles of N2H4 (limiting reactant) = 2 × 0.624 = 1.248 moles

5. Convert moles of products to mass

Now we will convert the moles of products to mass using their respective molar masses. Molar mass of N2 = (2 × 14.01 g/mol) = 28.02 g/mol Mass of N2 = (0.624 moles) × (28.02 g/mol) = 17.49 g Molar mass of H2O = (2 × 1.01 g/mol (H) + 16.00 g/mol (O)) = 18.02 g/mol Mass of H2O = (1.248 moles) × (18.02 g/mol) = 22.50 g In conclusion, when 20.0 g of pure hydrazine is ignited in the presence of 20.0 g of pure oxygen, 0.624 moles (17.49 g) of nitrogen and 1.248 moles (22.50 g) of water are produced as gaseous products.

Key concepts

Understanding Chemical Reactions

Chemical reactions involve substances called reactants transforming into different substances known as products. In the given reaction: \( \text{N}_2\text{H}_4 (l) + \text{O}_2 (g) \rightarrow \text{N}_2 (g) + 2 \text{H}_2\text{O} (g) \), hydrazine \((\text{N}_2\text{H}_4)\) reacts with oxygen \((\text{O}_2)\) to form nitrogen \((\text{N}_2)\) and water \((\text{H}_2\text{O})\). This process releases energy, making it an exothermic reaction suitable for rocket fuel.

Identifying the Limiting Reactant

In stoichiometry, the limiting reactant is the one that gets used up first in a chemical reaction, limiting the amount of products formed. To identify it, we compare the mole ratio of reactants. For our reaction, \( 0.624 \) moles of hydrazine reacts with \( 0.625 \) moles of oxygen. According to the balanced equation, these substances react in a 1:1 ratio. Since we have slightly less hydrazine than oxygen, hydrazine is the limiting reactant. This determines the maximum amount of products possible.

Calculating Molar Mass

Molar mass is the mass of one mole of a substance. It's crucial for converting between grams and moles, a common step in stoichiometry.
- For hydrazine: \( \text{N}_2\text{H}_4 \) has a molar mass of \( 32.05 \) g/mol (calculated as \( 2 \times 14.01 + 4 \times 1.01 \)).
- For oxygen: \( \text{O}_2 \) has a molar mass of \( 32.00 \) g/mol.
This conversion helps in determining the amounts used and ultimately the products formed.

Understanding Product Yield

Product yield refers to the quantity of product formed in a chemical reaction. It can be expressed in moles or grams. Using stoichiometry, the yield can be calculated based on the limiting reactant. In our exercise,

• Nitrogen yield: \( 0.624 \) moles equals \( 17.49 \) grams
• Water yield: \( 1.248 \) moles equals \( 22.50 \) grams

These calculations show how much product can actually be obtained from given reactants.