If 5.00 grams of hydrogen gas (H2) react with 25.0 grams of oxygen gas (O2) to form water (H2O), which is the limiting reactant? Calculate the theoretical yield of water in grams and the percent yield if the actual yield is 4.50 grams.
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To determine the limiting reactant, we need to compare the moles of hydrogen gas and oxygen gas.
1. Calculate the number of moles of hydrogen gas (H2):
5.00 g H2 * (1 mol H2 / 2.02 g H2) = 2.48 mol H2
2. Calculate the number of moles of oxygen gas (O2):
25.0 g O2 * (1 mol O2 / 32.00 g O2) = 0.78 mol O2
From the balanced equation: 2 H2 + O2 -> 2 H2O, we can see that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O.
3. Determine the ratio of moles of H2 to moles of O2:
2.48 mol H2 / 0.78 mol O2 ≈ 3.179
This means that for every 3.179 moles of H2, we need 1 mole of O2 for complete reaction. Since we can't have partial moles of a reactant, it means that oxygen gas (O2) is the limiting reactant.
4. Calculate the theoretical yield of water (H2O) using the limiting reactant:
0.78 mol O2 * (2 mol H2O / 1 mol O2) * (18.02 g H2O / 1 mol H2O) = 21.25 g H2O (theoretical yield)
5. Calculate the percent yield:
Percent yield = (actual yield / theoretical yield) * 100
Percent yield = (4.50 g / 21.25 g) * 100 ≈ 21.2%
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