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How many grams of NH3 can be produced from 4.10 moles of N and excess H?
Wiki User
∙ 6y ago
There is no compound NH. However, there is ammonia, NH3. The reactants are nitrogen gas, N2, and hydrogen gas, H2.
Multiply moles N2 by the mole ratio from the balanced equation between NH3 and N2, so that NH3 is in the numerator. Then multiply by the molar mass of NH3, 17.031 g/mol.
Balanced equation: N2 + 3H2 --> 2NH3
4.10 mol N2 x (2 mol NH3)/(1 mol N2) x (17.031 g NH3)/(1 mol NH3) = 140. g NH3 = 1.40 x 10^2 g NH3 rounded to three significant figures.
Wiki User
∙ 6y ago
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