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?

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 = per minute

                  = 1.116 moles per minute

No. of moles of being fed per minute = per minute

                 = 0.714 moles per minute

As, the balanced reaction equation is as follows.

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

Or,          0.71 moles of contains

                            = 0.357 moles of methanol.

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

Observed yield of methanol = 5.30 gram per minute

                                         =

                                          = 0.165 moles per minute

Nowe, we will calculate the percentage yield as follows.

       % yield =

                      =

                      = 46.32%

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

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.