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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.

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 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 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.