For the reaction n2 (g) + 2h2 (g) â n2h4 (l), if the percent yield for this reaction is 77.5%, what is the actual mass of hydrazine (n2h4) produced when 27.70 g of nitrogen reacts with 4.45 g of hydrogen?

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Chemistry

Maximilian

30 December, 08:55

For the reaction n2 (g) + 2h2 (g) â n2h4 (l), if the percent yield for this reaction is 77.5%, what is the actual mass of hydrazine (n2h4) produced when 27.70 g of nitrogen reacts with 4.45 g of hydrogen?

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Maritza · Answer 1

First calculate the moles of N2 and H2 reacted.

moles N2 = 27.7 g / (28 g/mol) = 0.9893 mol

moles H2 = 4.45 g / (2 g/mol) = 2.225 mol

We can see that N2 is the limiting reactant, therefore we base our calculation from that.

Calculating for mass of N2H4 formed:

mass N2H4 = 0.9893 mol N2 * (1 mole N2H4 / 1 mole N2) * 32 g / mol * 0.775

mass N2H4 = 24.53 grams