Liquid octane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. If 1.75g of water is produced from the reaction of 4.6g of octane and 4.8g of oxygen gas, calculate the percent yield of water... ___%?
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To determine the percent yield of water, we need to compare the actual yield (1.75g) to the theoretical yield of water.
First, let's calculate the moles of octane (C8H18) and oxygen gas (O2) used in the reaction:
Molar mass of octane (C8H18) = 8(12.01 g/mol) + 18(1.01 g/mol) = 114.23 g/mol
Molar mass of oxygen gas (O2) = 2(16.00 g/mol) = 32.00 g/mol
Number of moles of octane = mass/molar mass = 4.6 g / 114.23 g/mol = 0.0402 mol
Number of moles of oxygen gas = mass/molar mass = 4.8 g / 32.00 g/mol = 0.15 mol
Now, let's determine the mole ratio between octane and water. From the balanced chemical equation, we know that the ratio is 2:18 (2 moles of octane react to produce 18 moles of water).
Therefore, the moles of water produced would be:
Moles of water = (0.0402 mol octane) × (18 mol water / 2 mol octane) = 0.3624 mol
Next, we can calculate the theoretical yield of water using the molar mass of water (H2O):
Molar mass of water (H2O) = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Theoretical yield of water = moles of water × molar mass of water = 0.3624 mol × 18.02 g/mol = 6.53 g
Now we can calculate the percent yield of water using the actual yield and theoretical yield:
Percent yield = (Actual yield / Theoretical yield) × 100
Percent yield = (1.75 g / 6.53 g) × 100 = 26.8%
Therefore, the percent yield of water in this reaction is 26.8%.