Question

The reaction of hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) with hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) has been used in rocket engines. One way these compounds react is described by the equation $$ \mathrm{N}_{2} \mathrm{H}_{4}+7 \mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{HNO}_{3}+8 \mathrm{H}_{2} \mathrm{O} $$ According to this equation, how many grams of \(\mathrm{H}_{2} \mathrm{O}_{2}\) are needed to react completely with \(852 \mathrm{~g}\) of \(\mathrm{N}_{2} \mathrm{H}_{4} ?\)

Short answer

6324.38 grams of \(\mathrm{H}_{2} \mathrm{O}_{2}\) are needed to react completely with 852 grams of \(\mathrm{N}_{2} \mathrm{H}_{4}\).

Step-by-step solution

Calculate Molar Mass of Hydrazine

Find the molar mass of \(\mathrm{N}_{2} \mathrm{ H}_{4}\) by adding the atomic masses of nitrogen and hydrogen. The atomic mass of nitrogen (\(\mathrm{N}\)) is approximately 14.01 g/mol, and the atomic mass of hydrogen (\(\mathrm{H}\)) is approximately 1.01 g/mol. Thus, the molar mass of \(\mathrm{N}_{2} \mathrm{ H}_{4}\) is \(2 \times 14.01 + 4 \times 1.01 = 32.06 \) g/mol.

Calculate Moles of Hydrazine

To find out how many moles of \(\mathrm{N}_{2} \mathrm{H}_{4}\) are in 852 grams, use the formula: moles = mass / molar mass. Calculate the moles of \(\mathrm{N}_{2} \mathrm{H}_{4}\): \(\frac{852 \text{ g}}{32.06 \text{ g/mol}}\approx 26.56 \text{mol}\).

Determine Moles of Hydrogen Peroxide Needed

From the balanced chemical equation, 1 mole of \(\mathrm{N}_{2} \mathrm{ H}_{4}\) reacts with 7 moles of \(\mathrm{H}_{2} \mathrm{O}_{2}\). Therefore, for 26.56 moles of \(\mathrm{N}_{2} \mathrm{ H}_{4}\), multiply by 7 to get the moles of \(\mathrm{H}_{2} \mathrm{O}_{2}\) needed: \(26.56 \text{mol} \times 7 = 185.92 \text{mol}\).

Calculate Molar Mass of Hydrogen Peroxide

Find the molar mass of \(\mathrm{H}_{2} \mathrm{O}_{2}\) by adding the atomic masses of hydrogen and oxygen. The atomic mass of oxygen is approximately 16.00 g/mol. Thus, the molar mass of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(2 \times 1.01 + 2 \times 16.00 = 34.02 \) g/mol.

Calculate Mass of Hydrogen Peroxide Needed

To find the mass of \(\mathrm{H}_{2} \mathrm{O}_{2}\) needed to react with 852 g of \(\mathrm{N}_{2} \mathrm{H}_{4}\), use the formula: mass = moles \times molar mass. Calculate the mass of \(\mathrm{H}_{2} \mathrm{O}_{2}\) needed: \(185.92 \text{mol} \times 34.02 \text{g/mol} \approx 6324.38 \text{g}\).

Key concepts

Molar Mass Calculation

Understanding molar mass is crucial when dealing with chemical reactions. Molar mass, often expressed in grams per mole (g/mol), is the weight of one mole of a substance. It is calculated by adding up the atomic masses of all the atoms in a molecule.

For example, let's consider hydrazine ((N_2H_4)). Each molecule of hydrazine has two nitrogen atoms and four hydrogen atoms. To calculate its molar mass, you would multiply the atomic mass of nitrogen (14.01 g/mol) by two and the atomic mass of hydrogen (1.01 g/mol) by four, and then sum those numbers. This gives a molar mass of 32.06 g/mol for hydrazine.

It is important to note that atomic masses are averaged values based on isotopic distribution and are thus approximate. However, they are sufficiently accurate for most stoichiometric calculations in chemistry.

Chemical Reaction Balancing

Balancing chemical equations is a foundational skill in chemistry. It ensures that the law of conservation of mass is respected, meaning the number of atoms for each element is the same on both sides of the equation.

Consider the reaction between hydrazine ((N_2H_4)) and hydrogen peroxide ((H_2O_2)). The unbalanced equation is simple to write down as you just need to list the reactants and the products. However, to balance it, you must adjust the coefficients, the numbers in front of the compounds, so that the number of each type of atom is the same on both sides.

In the given example, the coefficient for hydrogen peroxide is seven. This ensures that for every one molecule of hydrazine, seven molecules of hydrogen peroxide are consumed, resulting in a balanced reaction showcasing an integral part of stoichiometry.

Mole-to-Mass Conversion

Once you have a balanced chemical equation, converting moles to mass is a straightforward process but pivotal for comprehending the amount of substances involved in a reaction. This process helps in determining how much of each reactant is needed and what amount of product will be formed.

In our example, we’re given the mass of hydrazine and asked to find the mass of hydrogen peroxide needed to react with it. By first converting the mass of hydrazine to moles, we can then use the stoichiometry of the balanced equation to find the corresponding moles of hydrogen peroxide. Finally, by knowing the molar mass of hydrogen peroxide, we can convert these moles back to grams to find the required mass. This mole-to-mass conversion is crucial in practical applications like determining the amount of fuel needed in rocket engines.