The maximum amount of n2h4 (hydrazine) that can be prepared from 14 grams of nitrogen and 28 grams of hydrogen by the reaction n2 + 2 h2 = n2h4 is?

Question

Chemistry

Ashlee English

8 November, 09:01

The maximum amount of n2h4 (hydrazine) that can be prepared from 14 grams of nitrogen and 28 grams of hydrogen by the reaction n2 + 2 h2 = n2h4 is?

+3

Answers (1)

Know the Answer?

Wilcox · Answer 1

To find out the maximum amount of N2H4 that can be prepared, we must determine which reactant is the controlling factor first. Controlling factor is the reactant that is converted completely during the reaction. Therefore we must first find out the moles of the each reactant (mass/molar mass).

Moles of N2 = 14 g / (28 g/mole) = 0.5 mole

Moles of H2 = 28 g / (2 g/mole) = 14 mole

Because the reaction ratio between N2 and H2 is 1:2, for a complete reaction of 0.5 moles N2, 1 mole H2 is needed. Clearly there are excess amount H2 present. So N2 is the controlling factor.

The ratio between N2 and N2H4 is 1:1, so the maximum moles of N2H4 generated from the reaction is also 0.5 mole. The maximum mass of N2H4 is 0.5 mole * 32 g/mole = 16 g.