Methanol CH3Oh can be produced by reacting carbon monoxide and hydrogen gas in the presence of a catalyst. If 75.00 grams of carbon monoxide reacts with 106.3 grams of hydrogen, 68.40 grams of methanol are produced. What is the percent yield of methanol?
can someone help me set this problem up.
what is the theoretical yield
Read my response. Next to last sentence.
"Now convert 3 mols CH3OH to grams. g = mols x molar mass. This is the theoretical yield (that is 100% yield)."
To determine the percent yield of methanol in this reaction, we need to compare the actual amount of methanol produced (given as 68.40 grams) to the theoretical amount of methanol that could have been produced, based on the reactants used.
To set up the problem, we can follow these steps:
Step 1: Write and balance the chemical equation for the reaction.
From the given information, we can write the balanced chemical equation as:
CO + 2H2 → CH3OH
Step 2: Calculate the molar mass of carbon monoxide (CO), hydrogen gas (H2), and methanol (CH3OH).
The molar mass of carbon monoxide (CO) = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
The molar mass of hydrogen gas (H2) = 1.01 g/mol + 1.01 g/mol = 2.02 g/mol
The molar mass of methanol (CH3OH) = 12.01 g/mol + (3 × 1.01 g/mol) + 16.00 g/mol + 1.01 g/mol = 32.04 g/mol
Step 3: Convert the given masses of reactants and product to moles.
75.00 g CO × (1 mol CO / 28.01 g CO) = 2.68 mol CO
106.3 g H2 × (1 mol H2 / 2.02 g H2) = 52.71 mol H2
68.40 g CH3OH × (1 mol CH3OH / 32.04 g CH3OH) = 2.13 mol CH3OH
Step 4: Determine the limiting reactant.
To identify the limiting reactant, compare the moles of CO and H2. We can see that the mole ratio between CO and H2 is 1:2. Thus, for every 1 mol of CO, we need 2 mol of H2, which means we would need 2 * 2.68 = 5.36 mol of H2 to react completely. Since we only have 52.71 mol of H2, it is the limiting reactant.
Step 5: Calculate the theoretical yield of methanol.
Using the balanced equation, we know that 1 mol of CO reacts to produce 1 mol of CH3OH. Since we have found that CO is the limiting reactant, we can equate the number of moles of methanol formed to moles of CO used:
2.68 mol CO × (1 mol CH3OH / 1 mol CO) = 2.68 mol CH3OH
Step 6: Calculate the percentage yield.
Percentage yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100%:
Percentage yield = (actual yield / theoretical yield) × 100%
Percentage yield = (68.40 g / 2.68 mol) × 100%
Percentage yield = 25.52%
Therefore, the percent yield of methanol in this reaction is 25.52%.
Which is the limiting reagent?
CO + 2H2 ==> CH3OH
mols CO = gram/molar mass = approx 3
mols H2 = approx 53
Convert mols CO to mols CH3OH using the coefficients in the balanced equation. That's 3 x (1 mol CH3OH/1 mol CO) = approx 3
Do the same for H2 to CH3OH. That's
53 mol H2 x (1 mol CH3OH/2 mol H2) = appxox 26
You note you have two values for mols CH3OH and both can't be right. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent responsible for that number is the limiting reagent. So all of the CO will be used, you will have an excess of H2 remaining unreacted, and you will produce approx 3 mols CH3OH.
Now convert 3 mols CH3OH to grams. g = mols x molar mass. This is the theoretical yield (that is 100% yield). The actual yield in the problem is given as 68.4g
%yield = (actual yield/theor yield)*100 = ?