Ammonia, NH3, can be synthesized by the following balanced reaction: 2NO(g) + 5 H2(g)---> 2NH3 (g)+2H2O(g). Starting with77.5 g of NO and 30.0 g of H2, identify the limiting reagent and find the theoretical yield of ammonia
Question
Ammonia, NH3, can be synthesized by the following balanced reaction: 2NO(g) + 5 H2(g)---> 2NH3 (g)+2H2O(g). Starting with77.5 g of NO and 30.0 g of H2, identify the limiting reagent and find the theoretical yield of ammonia
Solution
First, we need to determine the number of moles for each reactant. We do this by dividing the given mass by the molar mass of each compound.
The molar mass of NO is approximately 30 g/mol and the molar mass of H2 is approximately 2 g/mol.
So, for NO: 77.5 g / 30 g/mol = 2.58 mol And for H2: 30.0 g / 2 g/mol = 15 mol
The balanced chemical equation tells us that the reaction consumes NO and H2 in a ratio of 2:5. So, for every 2 moles of NO, we need 5 moles of H2.
However, we have 2.58 moles of NO and 15 moles of H2. If we divide the number of moles of each reactant by the number of moles required (according to the balanced equation), we can see which reactant is limiting.
For NO: 2.58 mol / 2 = 1.29 For H2: 15 mol / 5 = 3
The smaller number indicates the limiting reactant. So, NO is the limiting reactant.
The balanced equation also tells us that 2 moles of NO will produce 2 moles of NH3. So, the theoretical yield of NH3 is the same as the number of moles of the limiting reactant, NO.
Therefore, the theoretical yield of NH3 is 2.58 moles. To convert this to grams, we multiply by the molar mass of NH3 (approximately 17 g/mol).
So, the theoretical yield of NH3 is 2.58 mol * 17 g/mol = 43.86 g.
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