b.
Methanol, CH3OH, which can be used as a fuel, can be formed by the reaction between carbon monoxide and hydrogen as depicted in the following balanced equation. 2H2 + CO �¨ CH3OH Suppose 356 g of CO and 65 g of H2 are mixed and allowed to react.
(i)
Which species is the limiting factor in the overall reaction? Explain the reason for your answer.
(ii)
Calculate the amount of methanol that could, theoretically, be produced.
(iii)
If 352 g of methanol was actually produced, what is the percentage yield for your reaction?
(i) Convert the masses of reactants to moles. 356 g of CO is 356/28 = 12.7 moles. 65 g of H2 is 65/2.016 = 32.2 moles. There are MORE than twice as many moles of H2, compared to CO, so CO is the limiting reactant.
(ii) The number of moles of CH3OH that can be produced is 12.7 . That equals the number of moles of CO available. That mass of CH3OH formed is
12.7*(32.03) = 406.8 g
(iii) 352/406.8 = ___ yield
(convert to %)
(i) Convert the masses of reactants to moles. 356 g of CO is 356/28 = 12.7 moles. 65 g of H2 is 65/2.016 = 32.2 moles. There are MORE than twice as many moles of H2, compared to CO, so CO is the limiting reactant.
(ii) The number of moles of CH3OH that can be produced is 12.7 . That equals the number of moles of CO available. That mass of CH3OH formed is
12.7*(32.03) = 406.8 g
(iii) 352/406.8 = ___ yield
(convert to %)
(i) Well, the limiting factor in this reaction is like the VIP of the party. It's the species that gets used up first and puts a stop to the party. In this case, we need to compare the number of moles of CO and H2 that we have.
To do that, we need to convert grams into moles using the molar mass. So, for CO, the molar mass is 28.01 g/mol. Dividing 356 g of CO by the molar mass gives us 12.71 mol.
For H2, the molar mass is 2.02 g/mol. Dividing 65 g of H2 by the molar mass gives us 32.18 mol.
Since the balanced equation is telling us that it takes 1 mol of CO to react with 2 mol of H2, we can see that we have more than enough H2 to react with the CO. Therefore, the limiting factor is CO.
(ii) Now, let's calculate the theoretical amount of methanol that could be produced. According to the balanced equation, 1 mol of CO reacts with 2 mol of H2 to produce 1 mol of CH3OH.
Since we have 12.71 mol of CO, we will end up with 12.71 mol of CH3OH.
To convert this into grams, we use the molar mass of methanol, which is 32.04 g/mol. Multiplying the mol by the molar mass, we get 407.43 g of methanol.
Therefore, the theoretical amount of methanol that could be produced is 407.43 g.
(iii) Lastly, let's calculate the percentage yield using the actual amounts. We were told that 352 g of methanol was produced.
The percentage yield is calculated using the formula: (actual yield / theoretical yield) x 100.
Plugging in the values, we get (352 g / 407.43 g) x 100 = 86.4%.
So, the percentage yield for the reaction is 86.4%. Better than average, but hey, this reaction could still use some improvement!
(i) To determine the limiting factor in the reaction, we need to compare the number of moles of each reactant. First, let's calculate the number of moles for each reactant.
The molar mass of carbon monoxide (CO) is 28 g/mol, and we have 356 g of CO.
Number of moles of CO = Mass of CO / Molar mass of CO
Number of moles of CO = 356 g / 28 g/mol
Number of moles of CO = 12.71 mol
The molar mass of hydrogen gas (H2) is 2 g/mol, and we have 65 g of H2.
Number of moles of H2 = Mass of H2 / Molar mass of H2
Number of moles of H2 = 65 g / 2 g/mol
Number of moles of H2 = 32.5 mol
According to the balanced equation, the stoichiometric ratio between CO and H2 is 1:2. This means that 1 mole of CO reacts with 2 moles of H2.
In this case, we have an excess of H2 since we have 32.5 moles of H2, which is more than enough to react with the available CO. Therefore, hydrogen gas (H2) is the limiting factor in the overall reaction.
(ii) Now, let's calculate the maximum amount of methanol (CH3OH) that can be produced using the limiting reactant.
Based on the balanced equation, 2 moles of H2 react with 1 mole of CH3OH. Therefore, the stoichiometric ratio is 2:1.
Number of moles of CH3OH = Number of moles of H2 / 2
Number of moles of CH3OH = 32.5 mol / 2
Number of moles of CH3OH = 16.25 mol
The molar mass of methanol is 32 g/mol.
Mass of methanol = Number of moles of CH3OH x Molar mass of CH3OH
Mass of methanol = 16.25 mol x 32 g/mol
Mass of methanol = 520 g
Therefore, theoretically, a maximum of 520 g of methanol could be produced.
(iii) If 352 g of methanol was actually produced, we can calculate the percentage yield using the formula:
Percentage yield = (Actual yield / Theoretical yield) x 100
Actual yield = 352 g
Theoretical yield = 520 g (as calculated above)
Percentage yield = (352 g / 520 g) x 100
Percentage yield = 67.69%
Therefore, the percentage yield for the reaction is 67.69%.
(i) To determine the limiting factor in the overall reaction, we need to compare the number of moles of each reactant to the stoichiometric ratio of the balanced equation.
First, let's calculate the number of moles of CO and H2:
Molar mass of CO (carbon monoxide) = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
Number of moles of CO = mass / molar mass = 356 g / 28.01 g/mol = 12.7 mol
Molar mass of H2 (hydrogen gas) = 1.01 g/mol
Number of moles of H2 = mass / molar mass = 65 g / 1.01 g/mol = 64.4 mol
Now, let's compare the moles of both reactants to the stoichiometric ratio of the balanced equation: 2H2 + CO → CH3OH
From the balanced equation, we can see that 1 mol of CO reacts with 2 mol of H2 to produce 1 mol of CH3OH.
The ratio between moles of CO and H2 is 1:2. Since we have 12.7 mol of CO and 64.4 mol of H2, the ratio is approximately 1:5.1.
The limiting reactant is the one that is completely consumed first in the reaction. In this case, the CO is the limiting factor because there is not enough H2 to react with all of it.
(ii) To calculate the theoretical amount of methanol that could be produced, we need to determine the number of moles of the limiting reactant (CO) and use the stoichiometry of the balanced equation.
From the previous calculation, we found that the number of moles of CO is 12.7 mol.
According to the balanced equation, 1 mol of CO reacts to produce 1 mol of CH3OH.
Therefore, the theoretical amount of CH3OH that could be produced is also 12.7 mol.
To convert this to grams, we will use the molar mass of methanol:
Molar mass of CH3OH = 12.01 g/mol + 3(1.01 g/mol) + 16.00 g/mol = 32.04 g/mol
The mass of CH3OH that could theoretically be produced is:
Mass = number of moles × molar mass = 12.7 mol × 32.04 g/mol = 407 g
So, theoretically, 407 g of methanol could be produced.
(iii) The percentage yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100.
Given that the actual yield is 352 g and the theoretical yield is 407 g, we can calculate the percentage yield:
Percentage yield = (actual yield / theoretical yield) × 100
Percentage yield = (352 g / 407 g) × 100 ≈ 86.38%
Therefore, the percentage yield for the reaction is approximately 86.38%.