Part A: How many moles of NH3 can be produced from 21.0 mol of H2 and excess N2?Part B: How many grams of NH3 can be produced from 3.60 mol of N2 and excess H2?Part C: How many grams of H2 are needed to produce 11.76g of NH3?Part D: How many molecules (not moles) of NH3 are produced from 8.86 * 10^-4g of H2?
Concepts and reasonThe question is based on the concept of limiting reagent.Limiting reagent is a reagent in chemical reaction that gets totally consumed when chemical reaction is complete.
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FundamentalsThere are two methods to determine the limiting reagent. One method is to find the mole ratio and then compare the mole ratio of reactants that get used in the reaction. Other method is to calculate the grams of products that are produced from the given quantities of reactants.The mass of a substance can be calculated by using number of moles and molar mass of that substance as follows.m = nMHere, n is number of moles and M is molar mass of the substance.
Part A2.10 moles of H2reacted with excess of N2. So H2 is a limiting reagent. The balanced chemical reaction can be written as follows:
Part AThe moles of NH3 produced are 14 moles.First write the balanced chemical equation. Then, solve the problem using unitary method.
Part B2360 moles of N2reacted with excess of H2. So N2 is a limiting reagent. The balanced chemical reaction can be written as follows:
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Now, mass of ammonia is calculated by using formula as follow.
Part BThe mass of NH3 produced is 122.6 g.
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Part CWrite the balanced chemical reaction as follows:
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Part DMolecules of NH3 produced are
First write the balanced chemical equation. Then, solve the problem using unitary method. After finding the moles of hydrogen produced next step is to find the molecules of ammonia produced.