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

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FundamentalsThere room two approaches to determine the limiting reagent. One an approach is to discover the mole ratio and also then compare the mole proportion of reaction that get used in the reaction. Other technique is to calculate the grams of assets that are developed from the provided quantities the reactants.The mass of a substance have the right to be calculate by using number of moles and also molar mass of that substance together follows.m = nMHere, n is number of moles and M is molar fixed of the substance.

Part A2.10 moles of H2reacted v excess of N2. Therefore H2 is a limiting reagent. The well balanced chemical reaction have the right to be created as follows: Part AThe moles of NH3 developed are 14 moles.First create the well balanced chemical equation. Then, resolve the problem 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 can be written as follows: image.png767×118 12.5 KB
Now, mass of ammonia is calculate by making use of formula together follow. Part BThe fixed of NH3 developed is 122.6 g.

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Part CWrite the balanced chemical reaction together follows:    image.png694×107 15.6 KB

Part DMolecules the NH3 created are First compose the balanced chemical equation. Then, fix the trouble using unitary method. After finding the mole of hydrogen developed next step is to discover the molecule of ammonia produced.