in reaction S + O2 -> SO2, determine percent yield if 4 grams of S produced 64 gram of SO2
To determine the percent yield, we need to calculate the theoretical yield and the actual yield.
First, we need to find the molar mass of S (sulfur) and SO2 (sulfur dioxide). The molar masses are:
- Molar mass of S: 32 grams/mol
- Molar mass of SO2: 32 grams/mol (for sulfur) + 2 * 16 grams/mol (for oxygen) = 64 grams/mol
Next, we calculate the theoretical yield, which is the maximum amount of product that could be obtained if the reaction went to completion. We can use stoichiometry to determine the amount of SO2 that can be produced from the given amount of S.
Step 1: Convert the given mass of S to moles using its molar mass:
4 grams S * (1 mol S / 32 grams S) = 0.125 mol S
Step 2: Use the balanced equation to find the molar ratio between S and SO2:
1 mol S (from the balanced equation) produces 1 mol SO2 (from the balanced equation)
Step 3: Convert moles of S to moles of SO2:
0.125 mol S * (1 mol SO2 / 1 mol S) = 0.125 mol SO2
Step 4: Convert moles of SO2 to grams using its molar mass:
0.125 mol SO2 * (64 grams SO2 / 1 mol SO2) = 8 grams SO2 (theoretical yield)
Now, let's calculate the percent yield:
Percent yield = (Actual yield / Theoretical yield) * 100
The given information states that 4 grams of S produced 64 grams of SO2. Therefore, the actual yield is 64 grams.
Percent yield = (64 grams / 8 grams) * 100 = 800%
The percent yield is 800%.