Part A: How plenty of moles of NH3 have the right to be produced from 21.0 mol that H2 and excess N2?Part B: How plenty of grams of NH3 deserve to be created from 3.60 mol the N2 and also excess H2?Part C: How many grams that H2 are needed to create 11.76g that NH3?Part D: How plenty of molecules (not moles) the NH3 are created from 8.86 * 10^-4g that H2?

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

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FundamentalsThere space two approaches to recognize the limiting reagent. One technique is to uncover the mole ratio and then compare the mole proportion of reactants that get used in the reaction. Other an approach is to calculate the grams of assets that are produced from the offered quantities of reactants.The massive of a substance have the right to be calculation by using number of moles and molar fixed of that substance together follows.m = nMHere, n is variety of moles and also M is molar massive of the substance.

Part A2.10 moles of H2reacted v excess that N2. So H2 is a limiting reagent. The well balanced chemical reaction deserve to be written as follows:

Part AThe mole of NH3 produced are 14 moles.First create the balanced chemical equation. Then, resolve the difficulty using unitary method.

Part B2360 moles of N2reacted through excess the H2. For this reason N2 is a limiting reagent. The well balanced chemical reaction deserve to be written as follows:

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Now, massive of ammonia is calculation by making use of formula as follow.

Part BThe mass of NH3 created is 122.6 g.

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Part CWrite the well balanced chemical reaction as follows:

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Part DMolecules of NH3 created are

First write the balanced chemical equation. Then, resolve the difficulty using unitary method. After finding the mole of hydrogen produced next action is to uncover the molecule of ammonia produced.