Question

Methanol (CH3OH) can be produced by the following reaction: CO(g) 1 2H2(g) 88n CH3OH(g) 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 of methanol is produced per minute, what is the percent yield of the reaction?

Answer 1

Explanation:

It is know that, 1 mole of a gas at STP occupies a volume of 22. 4 L. Hence, calculate the number of moles of CO as follows.

     No. of moles of CO being fed per minute = $\frac{25.0 L}{22.4 L}$ per minute

                  = 1.116 moles per minute

No. of moles of $H_{2}$ being fed per minute = $\frac{16.0 L}{22.4 L}$ per minute

                 = 0.714 moles per minute

As, the balanced reaction equation is as follows.

          $CO + 2H_{2} \rightarrow CH_{3}OH$

Here, $H_{2}$ is the limiting reagent.  Since, 2 moles $H_{2}$ is equivalent to 1 mol methanol produced .

Or,          0.71 moles of $H_{2}$ contains $(\frac{1}{2}) \times 0.71 moles$

                            = 0.357 moles of methanol.

Therefore, theoretical yield of methanol = 0.357 moles per minute.

Observed yield of methanol = 5.30 gram per minute

                                         = $\frac{5.30 gram}{32.04 g/mol}$

                                          = 0.165 moles per minute

Nowe, we will calculate the percentage yield as follows.

       % yield = $\frac{\text{observed yield}}{\text{theoretical yield}} \times 100$

                      = $\frac{0.165 moles/min}{0.357 moles/min} \times 100$

                      = 46.32%

Thus, we can conclude that percent yield of the reaction is 46.32%.