Answer:
The percent yield of the reaction is 35 %
Explanation:
In the reaction, 1 mol of hydrazine reacts with 1 mol O₂ to produce 1 mol of nitrogen and 2 moles of water.
Let's verify the moles that were used in the reaction.
2.05 g . 1mol/ 32 g = 0.0640 mol
In the 100% yield, 1 mol of hydrazine produce 1 mol of N₂ so If I used 0.0640 moles of reactant, I made 0.0640 moles of products.
Let's use the Ideal Gases Law equation to find out the real moles of nitrogen, I made (real yield).
1atm . 0.550L = n . 0.082 . 295K
(1atm . 0.550L) / 0.082 . 295K = n → 0.0225 moles
Percent yield of reaction = (Real yield / Theoretical yield) . 100
(0.0225 / 0.0640) . 100 = 35%
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