Question

Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) a substance used as rocket fuel, reacts with oxygen as follows: $$\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ (a) Balance the equation. (b) How many moles of oxygen are needed to react with \(165 \mathrm{~g}\) of hydrazine? (c) How many grams of oxygen are needed to react with \(165 \mathrm{~g}\) of hydrazine?

Short answer

Balance equation: \(\mathrm{N}_{2} \mathrm{H}_{4} + 3 \mathrm{O}_{2} \rightarrow 2 \mathrm{NO}_{2} + 2 \mathrm{H}_{2} \mathrm{O}\). \(165\,\text{g}\,\mathrm{N}_2\mathrm{H}_4\) needs \(15.45\,\text{mol}\,\mathrm{O}_2\) or \(494.4\,\text{g}\).

Step-by-step solution

Balancing the Chemical Equation

The given chemical equation is:\[ \mathrm{N}_{2} \mathrm{H}_{4}(l) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(g) \]Start by balancing the nitrogen atoms: there are 2 N atoms in \(\mathrm{N}_2\mathrm{H}_4\), so we need 2 \(\mathrm{NO}_2\) on the right.\[ \mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2} + \mathrm{H}_{2} \mathrm{O} \]Next, balance the hydrogen: there are 4 H atoms in \(\mathrm{N}_2\mathrm{H}_4\), so we need 2 \(\mathrm{H}_2\mathrm{O}\) molecules.\[ \mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2} + 2 \mathrm{H}_{2} \mathrm{O} \]Now, balance the oxygen atoms: 4 O from \(2 \mathrm{NO}_{2}\) and 2 O from \(2 \mathrm{H}_2\mathrm{O}\), totaling 6 oxygen atoms required. So, 3 \(\mathrm{O}_2\) are needed.The balanced equation is:\[ \mathrm{N}_{2} \mathrm{H}_{4} + 3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2} + 2 \mathrm{H}_{2} \mathrm{O} \]

Calculate Moles of Hydrazine

The mass of hydrazine given is 165 g. The molar mass of \(\mathrm{N}_{2}\mathrm{H}_{4}\) can be calculated as follows:\[ \text{Molar mass of } \mathrm{N}_2\mathrm{H}_4 = 2(14.01) + 4(1.01) = 32.05 \, \text{g/mol} \]Calculate moles of hydrazine:\[ \text{Moles of } \mathrm{N}_{2}\mathrm{H}_{4} = \frac{165 \, \text{g}}{32.05 \, \text{g/mol}} = 5.15 \, \text{mol} \]

Calculate Moles of Oxygen

From the balanced equation, \(1\,\text{mol}\,\mathrm{N}_2\mathrm{H}_4\) reacts with \(3\,\text{mol}\,\mathrm{O}_2\). Therefore:\[ 5.15 \, \text{mol of } \mathrm{N}_{2}\mathrm{H}_{4} \times \frac{3 \, \text{mol } \mathrm{O}_{2}}{1 \, \text{mol } \mathrm{N}_{2}\mathrm{H}_{4}} = 15.45 \, \text{mol}\, \mathrm{O}_{2} \]

Calculate Mass of Oxygen Needed

The molar mass of \(\mathrm{O}_{2}\) is:\[ \text{Molar mass of } \mathrm{O}_2 = 2(16.00) = 32.00 \, \text{g/mol} \]Calculate the mass of oxygen needed using the moles calculated:\[ \text{Mass of } \mathrm{O}_2 = 15.45 \, \text{mol} \times 32.00 \, \text{g/mol} = 494.4 \, \text{g } \mathrm{O}_2 \]

Key concepts

Molar Mass Calculation

Understanding molar mass is like finding the weight of one mole of any substance. It is super important for converting between grams and moles, especially when dealing with chemical reactions. Molar mass is calculated by adding up the atomic masses of all atoms in a molecule. You can find these atomic masses on the periodic table.

For example, hydrazine \(\mathrm{N}_2\mathrm{H}_4\) consists of hydrogen (H) and nitrogen (N) atoms. The atomic mass of nitrogen is approximately 14.01 amu, and hydrogen is about 1.01 amu. If you sum up the mass of two nitrogen atoms (2 x 14.01) and four hydrogen atoms (4 x 1.01), the molar mass of hydrazine is 32.05 g/mol. Carrying out these calculations helps to find out how many moles are present in a given mass of a substance. This is important for predicting the outcomes of chemical reactions.

Stoichiometry

Stoichiometry is the math behind chemistry, allowing you to figure out how much of each substance is involved in a reaction. It is based on the balanced chemical equation, where the coefficients in the equation give the molar ratios of reactants and products.

In the reaction involving hydrazine and oxygen, the balanced equation is \ \mathrm{N}_{2} \mathrm{H}_{4} + 3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2} + 2 \mathrm{H}_{2} \mathrm{O} \ . This tells us that 1 mole of hydrazine reacts with 3 moles of oxygen to produce 2 moles of nitrogen dioxide and 2 moles of water.

If you have 5.15 moles of hydrazine, you can use the mole ratio from the balanced equation (1 mole of \(\mathrm{N}_2\mathrm{H}_4\) to 3 moles of \(\mathrm{O}_2\)) to find out how many moles of oxygen are required: \(5.15 \, \text{mol of } \mathrm{N}_{2}\mathrm{H}_{4} \times \frac{3 \, \text{mol } \mathrm{O}_{2}}{1 \, \text{mol } \mathrm{N}_{2}\mathrm{H}_{4}} = 15.45 \, \text{mol}\, \mathrm{O}_{2}\). This simple multiplication allows for accurate calculations of reactants and products.

Chemical Reactions

Chemical reactions involve the rearrangement of atoms to form new substances. Balancing chemical equations is essential because it ensures that the same number of each type of atom appears on both sides of the equation, following the Law of Conservation of Mass.

In the reaction of hydrazine with oxygen, the equation must be balanced to accurately reflect how the reactants convert to products. Balancing the equation \ \mathrm{N}_{2} \mathrm{H}_{4}(l) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(g) \ ensures that:

• 2 nitrogen atoms stay 2 nitrogen atoms in \(2 \mathrm{NO}_{2}\).
• 4 hydrogen atoms become 4 hydrogen atoms in \(2 \mathrm{H}_{2}\mathrm{O}\).
• 6 oxygen atoms (from 3 \(\mathrm{O}_2\)) are redistributed as 4 in 2 \(\mathrm{NO}_{2}\) and 2 in \(2 \mathrm{H}_2\mathrm{O}\).

This accurate equation allows chemists to predict how much product can be created from given reactants.