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11.6 (1) - the limiting reactant The limiting reactant...

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Key: 6. (Try this without a calculator) If I react 44.8 L of hydrogen gas with 56.0 g of nitrogen gas at STP, what is the theoretical yield of ammonia? a. 1.8 mol b. 4.5 mol c. 3.0 mol d. 1.3 mol e. 10.5 mol f. 7.8 mol g. none of these The problem asks you to find the theoretical yield of ammonia in the reaction described. Theoretical yield means the amount of product you get if everything reacts perfectly as opposed to the actual yield which you can only determine experimentally. To perform the dimensional analysis, you will need to create a balanced equation, which can be done from information in the problem: 3H 2 + N 2 = 2NH 3 In order to determine the amount of product formed, we need to know how much of the reactants reacted. Since one of the reactants will probably run out before the other, thus stopping the reaction, it is called the limiting reactant. Since the limiting reactant determines how much product is formed, we don’t care about the other reactant. The first step is therefore to determine
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Unformatted text preview: the limiting reactant. The limiting reactant could be either H 2 gas or N 2 gas. We could figure out which one is the limiting reactant by seeing which one would make less product if it were to react completely. This can be done using dimensional analysis. We us the balanced equation, 3 H 2 + N 2 = 2NH 3 56g N 2 │1mol N 2 │2mol NH 3 │ = 4mol NH 3 28g N 2 │1mol N 2 │ 44.8L H 2 │1mol H 2 │2mol NH 3 │ = 1.3mol NH 3 22.4L H 2 │3mol H 2 │ *note: don’t be confused because H 2 is given in liters. Remember that 1 mol of gas at STP occupies 22.4L. If you forget this, you can always plug in the standard temperature and pressure conditions (273K, and 1atm) into the ideal gas equation(PV = nRT). It looks like H 2 is the limiting reactant because it creates less product, and that it produces 1.3 moles of NH 3 ....
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