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: option (b) is the correct answer.

explanation:

a covalent bond is formed by sharing of electrons. whereas polarity arises due to difference in electronegativity of two atoms.

therefore, more is the electronegativity of an atom which combines with the titanium, more readily it will induce a partial positive and negative charge.

thus, out of the given options fluorine is most electronegative and has an incomplete octet. therefore, it will readily combine with titanium to share the electrons and complete its octet.  

1 is the answer to this problem