Here's a more detailed step-by-step calculation for the theoretical yield and percent yield of chromium (Cr) in the given reaction:
Given: Mass of chromium(III) oxide (Cr2O3) = 80.0 g Mass of carbon (C) = 8.00 g Actual yield of Cr = 21.7 g
Step 1: Calculate the molar mass of Cr2O3 and C. Molar mass of Cr2O3 = 2 x (51.996 g/mol) + 3 x (15.999 g/mol) = 151.996 g/mol Molar mass of C = 12.011 g/mol
Step 2: Convert the masses of Cr2O3 and C to moles. Moles of Cr2O3 = Mass of Cr2O3 / Molar mass of Cr2O3 = 80.0 g / 151.996 g/mol = 0.527 mol (rounded to three decimal places)
Moles of C = Mass of C / Molar mass of C = 8.00 g / 12.011 g/mol = 0.666 mol (rounded to three decimal places)
Step 3: Determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed. In this case, we compare the moles of Cr2O3 and C to see which one is limiting.
From the balanced equation: Cr2O3 + 3C -> 2Cr + 3CO
We can see that 1 mol of Cr2O3 requires 3 moles of C to react completely and produce 2 moles of Cr. Therefore, the limiting reactant is C, as we have 0.666 mol of C, which is less than the 0.527 mol of Cr2O3.
Step 4: Calculate the theoretical yield of Cr. The theoretical yield of Cr is the maximum amount of Cr that can be obtained based on the limiting reactant.
Moles of limiting reactant (C) = 0.666 mol Molar mass of Cr = 51.996 g/mol
Theoretical yield of Cr = Moles of limiting reactant (C) x Molar mass of Cr = 0.666 mol x 51.996 g/mol = 34.65 g (rounded to two decimal places)
Step 5: Calculate the percent yield of Cr. The percent yield is a measure of how much of the theoretical yield was actually obtained.
Actual yield of Cr = 21.7 g Theoretical yield of Cr = 34.65 g
Percent yield = (Actual yield / Theoretical yield) x 100% = (21.7 g / 34.65 g) x 100% = 62.7% (rounded to three significant figures)
Therefore, the percent yield for the reaction is approximately 62.7%.