Consider the reaction between hydrazine, N₂H₄, and hydrogen peroxide, H₂O₂, as shown in the unbalanced reaction below. This reaction has been used to propel rockets. __ N₂H₄ + __ H₂O₂→__ N₂ + __ H₂O (unbalanced reaction) If 20.0 grams of H₂O were isolated in a reaction starting with 10.0 grams of N₂H₄, what is the percentage yield for this reaction?

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To calculate the percentage yield, first, we need to balance the chemical equation and then determine the theoretical yield of water (H₂O) based on the starting amount of hydrazine (N₂H₄). The balanced chemical equation for the reaction between hydrazine (N₂H₄) and hydrogen peroxide (H₂O₂) is:$$ N_2H_4 + 2H_2O_2 \rightarrow N_2 + 4H_2O $$From the balanced equation, 1 mole of N₂H₄ produces 4 moles of H₂O. The molar mass of N₂H₄ is approximately 32.05 g/mol, and the molar mass of H₂O is approximately 18.015 g/mol.Given 10.0 grams of N₂H₄, the moles of N₂H₄ are:$$ \text{Moles of } N_2H_4 = \frac{10.0 \text{ g}}{32.05 \text{ g/mol}} \approx 0.312 \text{ moles} $$Since 1 mole of N₂H₄ produces 4 moles of H₂O, 0.312 moles of N₂H₄ will produce:$$ 0.312 \text{ moles} \times 4 = 1.248 \text{ moles of H}_2O $$The theoretical yield of H₂O in grams is:$$ 1.248 \text{ moles} \times 18.015 \text{ g/mol} \approx 22.5 \text{ grams} $$Given that 20.0 grams of H₂O were actually produced, the percentage yield is:$$ \text{Percentage yield} = \left( \frac{20.0 \text{ g}}{22.5 \text{ g}} \right) \times 100\% \approx 88.9\% $$

Consider the Reaction Between Hydrazine, N2H4, and Hydrogen Peroxide, H2O2

Consider the Reaction Between Hydrazine, N₂H₄, and Hydrogen Peroxide, H₂O₂