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Multiple Choice
Ammonia, NH3, reacts with hypochlorite ion, OCl–, to produce hydrazine, N2H4. How many grams of hydrazine are produced from 115.0 g NH3 if the reaction has a 81.5% yield? 2 NH3 + OCl– → N2H4 + Cl– + H2O
A
8818.3 g
B
10.82 g
C
108.2 g
D
88.2 g
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Verified step by step guidance
1
First, write the balanced chemical equation for the reaction: 2 NH3 + OCl^- → N2H4 + Cl^- + H2O.
Calculate the molar mass of NH3 (ammonia) by adding the atomic masses of nitrogen and hydrogen: Molar mass of NH3 = 14.01 g/mol (N) + 3 * 1.01 g/mol (H).
Determine the number of moles of NH3 by dividing the given mass (115.0 g) by the molar mass of NH3.
Use the stoichiometry of the balanced equation to find the moles of N2H4 produced. According to the equation, 2 moles of NH3 produce 1 mole of N2H4.
Calculate the theoretical mass of N2H4 by multiplying the moles of N2H4 by its molar mass (32.05 g/mol). Then, account for the reaction yield by multiplying the theoretical mass by 81.5% (or 0.815) to find the actual mass of hydrazine produced.