I have this question and I am pretty sure on what to use to solve it but I am just stuck. The reaction of 8 g of hydrogen and 28 g of carbon monoxide gave 16 g of methanol. What was the theoretical yield if the percent yield was 50%?
actual yield
________________ X 100% = percent yield
theoretical yield
Any help or explantion of the process would be greatly appreciated.
(Actual/theoretical)*100 = 50
(A/T) = 50/100 = 0.50
A = 0.5T or
T = A/0.5 = 2A
Theo = 2*16(from the problem) = 32 g.
To solve this problem, you need to first identify the theoretical yield, which is the amount of product that would be obtained if the reaction were to go to completion.
The balanced chemical equation for the reaction is given as:
CO + 2H2 → CH3OH
First, calculate the molar mass of each compound:
Molar mass of CO = 12 + 16 = 28 g/mol
Molar mass of H2 = 1 + 1 = 2 g/mol
Molar mass of CH3OH = 12 + 4 + 16 = 32 g/mol
Next, calculate the number of moles for each reactant and product:
Moles of CO: 28 g CO / 28 g/mol = 1 mol CO
Moles of H2: 8 g H2 / 2 g/mol = 4 mol H2
Moles of CH3OH: 16 g CH3OH / 32 g/mol = 0.5 mol CH3OH
From the balanced chemical equation, you can see that 1 mol of CO reacts with 2 mol of H2 to produce 1 mol of CH3OH. Therefore, the theoretical yield of CH3OH is equal to the number of moles of CO used:
Theoretical yield of CH3OH = 1 mol CH3OH
Now, you can use the formula for percent yield to calculate the actual yield:
percent yield = (actual yield / theoretical yield) * 100
The percent yield is given as 50%, so you can rearrange the formula to solve for the actual yield:
actual yield = (percent yield * theoretical yield) / 100
Substituting the given values into the formula:
actual yield = (50 * 1 mol CH3OH) / 100 = 0.5 mol CH3OH
Finally, you can convert the actual yield in moles to grams by multiplying by the molar mass:
actual yield = 0.5 mol CH3OH * 32 g/mol = 16 g CH3OH
Therefore, the theoretical yield of methanol is 1 mole and the actual yield is 16 grams.