Question

Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\), cmits a large quantity of energy when it reacts with oxygen, which has led to hydrazine's use as a fuel for rockets: $$ \mathrm{N}_{2} \mathrm{H}_{4}(I)+\mathrm{O}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ 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

When 20.0 grams of hydrazine is ignited in the presence of 20.0 grams of oxygen, \(0.624\ \text{mol}\) of nitrogen and \(1.248\ \text{mol}\) of water are produced. This corresponds to 17.5 grams of nitrogen and 22.5 grams of water.

Step-by-step solution

Balanced chemical equation

We are given the balanced chemical equation: \(\mathrm{N_{2}H_{4}(l) + O_{2}(g) \rightarrow N_{2}(g) + 2H_{2}O(g)}\)

Calculate the number of moles of reactants

First, we need to find the molar masses of hydrazine \(\mathrm{(N_{2}H_{4})}\) and oxygen \(\mathrm{(O_{2})}\). \[\text{Molar mass of N}_{2}\text{H}_{4} = 2\times 14.01\ \text{g/mol (N)} + 4\times 1.008\ \text{g/mol (H)} = 32.05\ \text{g/mol}\] \[\text{Molar mass of O}_{2} = 2\times 16.00\ \text{g/mol} = 32.00\ \text{g/mol}\] Now we can calculate the moles of each reactant: \[n(\mathrm{N_{2}H_{4}}) = \frac{20.0\ \text{g}}{32.05\ \text{g/mol}} = 0.624\ \text{mol}\] \[n(\mathrm{O}_{2}) = \frac{20.0\ \text{g}}{32.00\ \text{g/mol}} = 0.625\ \text{mol}\]

Determine the limiting reactant

To identify the limiting reactant, we need to look at the balanced equation and find the reactant with the lower mole ratio. \[n\left(\frac{\text {N}_{2}\text {H}_{4}}{\text{O}_{2}}\right)=\frac{n(\text{N}_{2}\text {H}_{4})}{n(\text{O}_{2})}=\frac{0.624}{0.625}=0.998\] Since the mole ratio of \(\text{N}_{2}\text {H}_{4}\) to \(\text{O}_{2}\) is less than 1, \(\text{N}_{2}\text {H}_{4}\) is the limiting reactant.

Use stoichiometry to find the number of moles of each product

Using the balanced equation and the moles of the limiting reactant, we can find the moles of each product: \[n(\text{N}_{2}) = n(\text{N}_{2}\text {H}_{4}) = 0.624\ \text{mol}\] \[n(\text{H}_{2}\text{O}) = 2\times n(\text{N}_{2}\text {H}_{4}) = 2\times 0.624\ \text{mol} = 1.248\ \text{mol}\]

Convert the moles of products to grams

Finally, convert the moles of each product to grams using their molar masses: \[\text{Mass of N}_{2} = n(\text{N}_{2}) \times \text{molar mass of N}_{2} = 0.624\ \text{mol} \times 28.02\ \text{g/mol} = 17.5\ \text{g}\] \[\text{Mass of H}_{2}\text{O} = n(\text{H}_{2}\text{O}) \times \text{molar mass of H}_{2}\text{O} = 1.248\ \text{mol} \times 18.02\ \text{g/mol} = 22.5\ \text{g}\] Therefore, when 20.0 grams of hydrazine is ignited in the presence of 20.0 grams of oxygen, 17.5 grams of nitrogen and 22.5 grams of water are produced.

Key concepts

Chemical Equations

Chemical equations are the backbone of understanding chemical reactions. They represent what occurs at the atomic level during a reaction. In chemical equations, reactants (starting substances) and products (substances formed) are separated by an arrow. For example, in the equation \(\mathrm{N_{2}H_{4}(l) + O_{2}(g) \rightarrow N_{2}(g) + 2H_{2}O(g)}\), Hydrazine \((\mathrm{N_2H_4})\) and oxygen \((\mathrm{O_2})\) are the reactants, while nitrogen \((\mathrm{N_2})\) and water \((\mathrm{H_2O})\) are the products.
Balancing chemical equations is crucial to reflect the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction. This means the number of atoms for each element involved in the reaction must be the same on both sides of the equation. In our example, we ensure that there are equal numbers of nitrogen, hydrogen, and oxygen atoms on both sides of the equation. This step helps correctly predict the quantities of reactants and products involved in a reaction.

Limiting Reactant

In stoichiometry, the limiting reactant is the substance that is entirely consumed first in a chemical reaction, limiting the amount of product that can be formed. To find the limiting reactant, compare the mole ratio of the reactants you have with the mole ratio in the balanced equation.
In the given problem, both hydrazine and oxygen are used as reactants. Start by calculating their moles: hydrazine has \(0.624\ \text{mol}\) and oxygen has \(0.625\ \text{mol}\). Look at the balanced chemical equation: it shows a 1:1 ratio of hydrazine to oxygen. Thus, theoretically, we would need the same amount of moles for complete reaction. However, since \(0.624\) moles of hydrazine are slightly less than \(0.625\) moles of oxygen, hydrazine becomes the limiting reactant.
This concept ensures that no extra calculations are made for producing products from an amount of reactants that do not exist, making it a vital skill for predicting chemical yields accurately.

Molar Mass Calculation

Molar mass is essential for converting grams of a substance to moles. It acts as a bridge in stoichiometric calculations between mass and number of moles, allowing you to perform reaction predictions correctly.
To calculate the molar mass of a compound, sum the atomic masses (from the periodic table) of all atoms in a molecule. For example, hydrazine \((\mathrm{N_2H_4})\) has two nitrogen atoms (each \(14.01\ \text{g/mol}\)) and four hydrogen atoms (each \(1.008\ \text{g/mol}\)) leading to a molar mass of \(32.05\ \text{g/mol}\). Similarly, oxygen \((\mathrm{O_2})\)'s molar mass is \(32.00\ \text{g/mol}\) as it consists of two oxygen atoms (each \(16.00\ \text{g/mol}\)).
Understanding these simple conversions is crucial for calculating how many moles of reactants will react and the moles of products formed, completely from the given mass of substances.