Part A: How countless moles that NH3 have the right to be developed from 21.0 mol that H2 and excess N2?Part B: How many grams of NH3 can be produced from 3.60 mol that N2 and excess H2?Part C: How numerous grams of H2 are essential to produce 11.76g of NH3?Part D: How numerous molecules (not moles) that NH3 are developed from 8.86 * 10^-4g of H2?

Concepts and reasonThe inquiry is based on the principle of limiting reagent.Limiting reagent is a reagent in chemistry reaction the gets totally consumed when chemical reaction is complete.

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FundamentalsThere space two methods to determine the limiting reagent. One an approach is to uncover the mole ratio and also then compare the mole ratio of reactants that gain used in the reaction. Other method is to calculate the grams of commodities that are developed from the offered quantities of reactants.The mass of a substance can be calculated by using number of moles and molar fixed of the substance together follows.m = nMHere, n is number of moles and also M is molar fixed of the substance.

Answer

Part A2.10 moles of H2reacted v excess the N2. So H2 is a limiting reagent. The balanced chemical reaction can be created as follows:

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Part AThe mole of NH3 produced are 14 moles.First compose the well balanced chemical equation. Then, fix the trouble using unitary method.

Part B2360 moles of N2reacted through excess of H2. Therefore N2 is a limiting reagent. The well balanced chemical reaction can be written as follows:

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Now, massive of ammonia is calculation by using formula as follow.
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Part BThe massive of NH3 developed is 122.6 g.

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Part CWrite the balanced chemical reaction as follows:

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Part DMolecules that NH3 created are

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First create the balanced chemical equation. Then, solve the trouble using unitary method. ~ finding the moles of hydrogen developed next step is to uncover the molecules of ammonia produced.