Answer:
320.23g of CH3OH.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CO(g) + 2H2(g)—> CH3OH(l)
Next, we shall determine the masses of CO and H2 that reacted and the mass of CH3OH produced from the balanced equation. This is illustrated below below:
Molar mass of CO = 12 + 16 = 28g/mol
Mass of CO from the balanced equation = 1 x 28 = 28g
Molar mass of H2 = 2x1 = 2g/mol
Mass of H2 from the balanced equation = 2 x 2= 4g
Molar mass of CH3OH = 12 + (3x1) + 16 + 1 = 32g/mol
Mass of CH3OH from the balanced equation = 1 x 32 = 32g
From the balanced equation above,
28g of CO reacted with 4g of H2 to produce 32g of CH3OH.
Next, we shall determine the the limiting reactant. This is illustrated below:
From the balanced equation above,
28g of CO reacted with 4g of H2.
280.2g of CO will react with =
(280.2 x 4)/28 = 40.03g of H2.
From the calculations made above, we can see that only 40.03g out of 50.5g of H2 is required to react completely with 280.2g of CO.
Therefore, CO is the limiting reactant and H2 is the excess reactant.
Finally, we shall determine the mass of methanol, CH3OH produced from the reaction.
In this case, the limiting reactant will be used because it will give the maximum yield of the reaction since all of it is used up in the reaction. The limiting reactant is CO and the mass of methanol, CH3OH produced can be obtained as follow:
From the balanced equation above,
28g of CO reacted to produce 32g of CH3OH.
Therefore, 280.2g of CO will react to produce = (280.2 x 32)/28 = 320.23g of CH3OH.
Therefore, 320.23g of CH3OH were produced from the reaction.
