Question

Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\), once used as a rocket propellant, reacts with oxygen: $$ \mathrm{N}_{2} \mathrm{H}_{4}+\mathrm{O}_{2} \longrightarrow \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O} $$ (a) How many grams of \(\mathrm{O}_{2}\) are needed to react with \(50.0 \mathrm{~g}\) of \(\mathrm{N}_{2} \mathrm{H}_{4} ?\) (b) How many grams of \(\mathrm{N}_{2}\) are obtained if the yield is \(85.5 \% ?\)

Short answer

49.92 g of O2 are needed, and 37.36 g of N2 are obtained.

Step-by-step solution

Calculate Molar Mass of N2H4

To begin, calculate the molar mass of hydrazine, \(\mathrm{N}_2\mathrm{H}_4\). It has 2 nitrogen atoms and 4 hydrogen atoms. - Molar mass of nitrogen (N): 14.01 g/mol- Molar mass of hydrogen (H): 1.01 g/molMolar mass of \(\mathrm{N}_2\mathrm{H}_4 = (2 \times 14.01) + (4 \times 1.01) = 28.02 + 4.04 = 32.06\, \text{g/mol}.\)

Convert Mass of N2H4 to Moles

Now, convert 50.0 g of \(\mathrm{N}_2\mathrm{H}_4\) to moles using its molar mass.\[\text{moles of } \mathrm{N}_2\mathrm{H}_4 = \frac{50.0\, \text{g}}{32.06\, \text{g/mol}} = 1.56\, \text{mol}.\]

Determine Moles of O2 Needed

From the balanced equation, \(1\) mole of \(\mathrm{N}_2\mathrm{H}_4\) reacts with \(1\) mole of \(\mathrm{O}_2\). Therefore, to react with \(1.56\, \text{mol}\) of \(\mathrm{N}_2\mathrm{H}_4\), we need \(1.56\, \text{mol}\) of \(\mathrm{O}_2\).

Convert Moles of O2 to Grams

Calculate the mass of \(\mathrm{O}_2\) required using its molar mass.- Molar mass of \(\mathrm{O}_2\) (Oxygen): \(32.00\, \text{g/mol}\).\[\text{mass of } \mathrm{O}_2 = 1.56\, \text{mol} \times 32.00\, \text{g/mol} = 49.92\, \text{g}.\]

Calculate Theoretical Yield of N2

Determine the theoretical yield of \(\mathrm{N}_2\). According to the balanced equation, \(1\) mole of \(\mathrm{N}_2\mathrm{H}_4\) produces \(1\) mole of \(\mathrm{N}_2\). The theoretical moles of \(\mathrm{N}_2\) = \(1.56\, \text{mol}\).

Convert Theoretical Moles of N2 to Grams

Use the molar mass of nitrogen (\(28.02\, \text{g/mol}\)) to find the mass.\[\text{mass of } \mathrm{N}_2 = 1.56\, \text{mol} \times 28.02\, \text{g/mol} = 43.70\, \text{g}.\]

Calculate Actual Yield of N2

The actual yield is given as \(85.5\%\) of the theoretical yield.\[\text{actual yield} = 43.70\, \text{g} \times 0.855 = 37.36\, \text{g}.\]

Key concepts

Molar Mass

Molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). To calculate the molar mass, sum the atomic masses of all atoms in a molecule. For example, in hydrazine, \(\mathrm{N}_2\mathrm{H}_4\), you have two nitrogen atoms and four hydrogen atoms.

• The atomic mass of nitrogen (N) is 14.01 g/mol.
• The atomic mass of hydrogen (H) is 1.01 g/mol.

Calculating molar mass involves these steps:

• Multiply the atomic mass of nitrogen by the number of nitrogen atoms: \(2 \times 14.01 = 28.02\ g/mol\).
• Multiply the atomic mass of hydrogen by the number of hydrogen atoms: \(4 \times 1.01 = 4.04\ g/mol\).
• Add these values together to find the molar mass of hydrazine: \(28.02 + 4.04 = 32.06\ g/mol.\)

Understanding molar mass is crucial for converting between grams and moles, a key step in stoichiometry.

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is completely consumed first, stopping the reaction. It determines the amount of products formed. To identify the limiting reactant, you compare the mole ratio of the reactants used in your balanced equation with what you actually have.

In our reaction, \(\mathrm{N}_2\mathrm{H}_4 + \mathrm{O}_2 \rightarrow \mathrm{N}_2 + 2\mathrm{H}_2\mathrm{O}\), we need one mole of \(\mathrm{O}_2\) for every mole of \(\mathrm{N}_2\mathrm{H}_4\). After converting the given mass of \(\mathrm{N}_2\mathrm{H}_4\) to moles, we find it requires the same moles of \(\mathrm{O}_2\).

Since the amount of \(\mathrm{N}_2\mathrm{H}_4\) directly tells how much \(\mathrm{O}_2\) is needed, you're comparing available reactants to their required molar quantities to find which runs out first. This process helps in maximizing and predicting product yield.

Theoretical Yield

Theoretical yield is the maximum amount of product that could be formed from the given quantities of reactants. It's calculated based on the balanced chemical equation and the limiting reactant. In other words, it assumes perfect conditions with no side reactions or losses.

• Calculate the moles of product expected from the reactant using stoichiometry.
• Using the molar mass, convert moles of product into grams to get the theoretical yield.

From the balanced equation, we found that 1 mole of \(\mathrm{N}_2\mathrm{H}_4\) yields 1 mole of \(\mathrm{N}_2\). Thus, we expect \(1.56\) moles of \(\mathrm{N}_2\) from \(50.0\) g of \(\mathrm{N}_2\mathrm{H}_4\).
Calculate the mass of this \(\mathrm{N}_2\) using its molar mass (28.02 g/mol), giving us the theoretical yield: 43.70 g.

It’s like the perfect score on a test—you strive to get there, but real conditions often mean less than perfect results.

Actual Yield

Actual yield is the real amount of product obtained from a reaction. It's often less than the theoretical yield due to losses or inefficiencies. To find the actual yield, you multiply the theoretical yield by the percent yield expressed as a decimal.

For example, if the theoretical yield of \(\mathrm{N}_2\) is 43.70 g, and the percent yield is 85.5%, the actual yield is calculated as follows:
\[ ext{actual yield} = 43.70 \, \text{g} \times 0.855 = 37.36 \, \text{g}.\]
The actual yield helps gauge the efficiency of the reaction and identify potential sources of product loss.