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eaction: co1g2 1 2h21g2 h ch3oh1g2 copyright 2018 cengage learning. all rights reserved. may not be copied, scanned, or duplicated, in whole or in part. wcn 02-200-203 hydrogen at stp flows into a reactor at a rate of 16.0 l/min. carbon monoxide at stp flows into the reactor at a rate of 25.0 l/min. if 5.30 g methanol is produced per minute, what is the percent yield of the reaction?

User Mlynn
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The percent yield of the reaction is 23.0%.

Here is the answer and explanation:

Percent Yield

The percent yield of a reaction is the actual yield of the product divided by the theoretical yield, multiplied by 100%. The theoretical yield is the amount of product that would be produced if the reaction went to completion.

We can calculate the molar flow rate using the following formula:

Molar flow rate = (flow rate) / (molar volume)

where:

Molar flow rate is in moles per minute (mol/min)

Flow rate is in liters per minute (L/min)

Molar volume is in liters per mole (L/mol)

The molar volume of a gas at STP is 22.4 L/mol. Using this information, we can calculate the molar flow rates of hydrogen and carbon monoxide:

Molar flow rate of H2 = (16.0 L/min) / (22.4 L/mol) = 0.72 mol/min

Molar flow rate of CO = (25.0 L/min) / (22.4 L/mol) = 1.11 mol/min

We can calculate the theoretical yield using the following formula:

Theoretical yield = (limiting reactant) * (moles of product per mole of limiting reactant)

where:

Theoretical yield is in moles

Limiting reactant is the reactant that is used up first

Moles of product per mole of limiting reactant is the stoichiometric coefficient of the product

The stoichiometric coefficient of methanol is 1, so the theoretical yield of methanol is:

Theoretical yield = (0.72 mol/min) * (1 mol/mol) = 0.72 mol/min

Finally, we can calculate the percent yield of the reaction:

Percent yield = (actual yield) / (theoretical yield) * 100%

where:

Percent yield is in percent (%)

Actual yield is in moles

Theoretical yield is in moles

The actual yield of methanol is 5.30 g/min. We can convert this to moles using the following formula:

Molar mass of methanol = 32.04 g/mol

Actual yield = (5.30 g/min) / (32.04 g/mol) = 0.165 mol/min

Using this information, we can calculate the percent yield of the reaction:

Percent yield = (0.165 mol/min) / (0.72 mol/min) * 100% = 23.0%

Therefore, the percent yield of the reaction is 23.0%.

Question

Reaction: co1g2 1 2h21g2 h ch3oh1g2 copyright 2018 cengage learning. all rights reserved. may not be copied, scanned, or duplicated, in whole or in part. wcn 02-200-203 hydrogen at stp flows into a reactor at a rate of 16.0 l/min. carbon monoxide at stp flows into the reactor at a rate of 25.0 l/min. if 5.30 g methanol is produced per minute, what is the percent yield of the reaction?

User NESHOM
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