Methyl alcohol (wood alcohol), CH3OH, is produced via the reaction
CO(g) + 2 H2(g) à CH3OH(l)
A mixture of 1.20 g H2(g) and 7.45 g CO(g) are allowed to react. 4 points
(a) Which reagent is the limiting reagent?
(b) What is the yield of CH3OH? [Assume theoretical yield in g is what is wanted here.]
(c) How much of the reagent present in excess is left over?
(d) Suppose the actual yield is 7.52 g of CH3OH. What is the % yield?
9.6
To determine the limiting reagent, we need to compare the moles of each reactant.
(a) Calculate the number of moles of each reactant:
The molar mass of H2 is 2 g/mol, and the molar mass of CO is 28 g/mol.
Number of moles of H2 = (1.20 g / 2 g/mol) = 0.6 mol
Number of moles of CO = (7.45 g / 28 g/mol) = 0.266 mol
The balanced equation tells us that the stoichiometric ratio between CO and H2 is 1:2. Therefore, the reactant with the smaller number of moles is the limiting reagent. In this case, CO has fewer moles, so it is the limiting reagent.
(b) To find the theoretical yield of CH3OH, we first need to find the number of moles of CO that reacted. From the balanced equation, we know that for every 1 mole of CO, we get 1 mole of CH3OH.
Number of moles of CH3OH = Number of moles of CO
Theoretical yield of CH3OH = (Number of moles of CH3OH) x (Molar mass of CH3OH)
The molar mass of CH3OH is 32 g/mol.
Theoretical yield of CH3OH = (0.266 mol) x (32 g/mol) = 8.512 g
(c) To find the amount of excess reagent left over, we need to determine the number of moles of the excess reagent and then convert it to grams. Since CO is the limiting reagent, the excess reagent is H2.
Number of moles of H2 remaining = Number of moles of H2 initially - Number of moles of H2 consumed
Number of moles of H2 remaining = 0.6 mol - (0.266 mol / 2) = 0.467 mol
Amount of H2 remaining = (Number of moles of H2 remaining) x (Molar mass of H2)
The molar mass of H2 is 2 g/mol.
Amount of H2 remaining = (0.467 mol) x (2 g/mol) = 0.934 g
(d) To find the percent yield, we divide the actual yield (given as 7.52 g) by the theoretical yield (calculated in part b), and then multiply by 100.
% Yield = (Actual yield / Theoretical yield) x 100
% Yield = (7.52 g / 8.512 g) x 100 = 88.5%
To determine the limiting reagent, you need to compare the number of moles of each reactant to the stoichiometry of the balanced equation. The reactant that produces the least amount of product is the limiting reagent.
(a) First, calculate the number of moles for each reactant:
Number of moles of H2 = mass of H2 / molar mass of H2
= 1.20 g / 2 g/mol
= 0.60 mol
Number of moles of CO = mass of CO / molar mass of CO
= 7.45 g / 28 g/mol
= 0.265 mol
Next, compare the ratios of the reactants to the stoichiometry of the balanced equation:
H2:CO = 0.60 mol : 0.265 mol
= 2.26 : 1
Since the ratio is not 1:1, we can see that H2 is the limiting reagent.
(b) The yield of CH3OH can be calculated based on the stoichiometry of the balanced equation. From the equation, 1 mol of CO produces 1 mol of CH3OH.
Number of moles of CH3OH = Number of moles of CO (since 1:1 ratio)
= 0.265 mol
Mass of CH3OH = number of moles of CH3OH x molar mass of CH3OH
= 0.265 mol x 32 g/mol
= 8.48 g
Therefore, the theoretical yield of CH3OH is 8.48 g.
(c) To determine the amount of excess reagent left over, you need to calculate the moles of the excess reagent consumed:
Moles of CO consumed = Moles of CH3OH produced (from the balanced equation)
= 0.265 mol
Moles of CO excess = Moles of CO initially - Moles of CO consumed
= 0.265 mol - 0.265 mol (since H2 is the limiting reagent)
= 0 mol
Therefore, there is no excess CO left over.
(d) The percent yield is calculated by dividing the actual yield by the theoretical yield, and multiplying by 100:
% Yield = (Actual yield / Theoretical yield) x 100
= (7.52 g / 8.48 g) x 100
= 88.7%
Therefore, the percent yield is 88.7%.