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Vladimir [108]
3 years ago
14

If 3.95 g of N2H4 reacts and produces 0.750 L of N2, at 295 K and 1.00 atm, what is the percent yield of the reaction?

Chemistry
1 answer:
adell [148]3 years ago
3 0
Molar mass of N2H4 = 32 grams/mole 
<span>3.95 grams of  N2H4 = 3.95/32
                                  = 0.123 moles </span>

<span>This will produce 0.123 moles of N2 </span>
<span>Now,
From the gas law equation. </span>
<span>P.V = n x R x T </span>
<span>P = 1 atm (given)
V = </span><span>0.123</span><span> x 0.082057 x 295 </span>
<span>V = 2.97 Liters </span>

<span>Theoretical yield = 2.97 Liters.
Actual yield = 0.750 Liters </span>

percentage yield = (0.75/2.97) x 100 %
                           = 25.25 %
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