For the reaction:
N2 + H2 --> NH3
what is the maximum amount of NH3 (17.0305 g/mol) which could be formed from 9.85 mol (28.0134 g/mol) of N2 and 4.11 mol of H2 (2.01588 g/mol)?
thank you so much!
N2 + H2 --> NH3
what is the maximum amount of NH3 (17.0305 g/mol) which could be formed from 9.85 mol (28.0134 g/mol) of N2 and 4.11 mol of H2 (2.01588 g/mol)?
thank you so much!
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This is a "limiting reactant" problem. There are two techniques,one is to use each amount of reactant as if it were the limiting reactant, and finding the amount of product. The actual amount of product formed is the smaller of the two. The second method is to determine the limiting reactant by comparing the amounts of reactants, and then find the amount of product using the limiting reactant.
Start by balancing the equation.
N2 + 3H2 --> 2NH3
You know that 9.85 mol of N2 reacts with three times that much hydrogen, but there is only 4.11 moles of hydrogen available, far less than the amount needed. Therefore, hydrogen is the limiting reactant, and the amount of ammonia produced will be 2/3 of 4.11 moles.
4.11 mol H2 x (2 mol NH3 / 3 mol H2) = 2.74 moles NH3 produced.
Start by balancing the equation.
N2 + 3H2 --> 2NH3
You know that 9.85 mol of N2 reacts with three times that much hydrogen, but there is only 4.11 moles of hydrogen available, far less than the amount needed. Therefore, hydrogen is the limiting reactant, and the amount of ammonia produced will be 2/3 of 4.11 moles.
4.11 mol H2 x (2 mol NH3 / 3 mol H2) = 2.74 moles NH3 produced.
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2.74 mol NH3 x (17.03 g NH3 / 1 mol NH3) = 46.7 g NH3
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