Question

Methanol can be produced through the reaction of CO and H2 in the presence of a catalyst.

CO(g) + 2H2(g) catalyst, CH;OH(0)
If 85 g of CO reacts to produce 61.8 g of
CH;OH, what is the percent yield of CH OH?
Answer in units of%.

Answer 1

63.6%

Step-by-step explanation:

To calculate the percent yield of CH3OH, we need to compare the actual yield to the theoretical yield, which is the maximum amount of CH3OH that can be produced based on the amount of CO and H2 that was used in the reaction.

The theoretical yield can be calculated using the balanced chemical equation for the reaction:

CO(g) + 2H2(g) -> CH3OH(l)

Since 1 mole of CO reacts with 2 moles of H2 to produce 1 mole of CH3OH, the moles of CH3OH produced can be calculated by dividing the mass of CO by its molar mass and then multiplying by the ratio of CH3OH to CO in the balanced equation (1 mole of CH3OH per 1 mole of CO).

Let's assume the moles of CO used in the reaction is n. Then,

n = 85 g CO / 28.01 g/mol = 3.03 moles

And the theoretical yield of CH3OH would be:

3.03 moles CO * (1 mole CH3OH/1 mole CO) = 3.03 moles CH3OH

The moles of CH3OH can then be converted to grams:

3.03 moles CH3OH * 32.04 g/mol = 97.36 g CH3OH

Finally, the percent yield can be calculated by dividing the actual yield of CH3OH (61.8 g) by the theoretical yield (97.36 g) and multiplying by 100%.

Percent yield = (Actual yield / Theoretical yield) * 100%

= (61.8 g CH3OH / 97.36 g CH3OH) * 100%

= 63.6%

So the percent yield of CH3OH is 63.6%.

tell me if you confus

ALLEN