After distilling your crude methyl benzoate, you set aside 4.83 grams of the purified ester. you then prepare the grignard reagent ( phenylmagnesium bromide ) by reacting 2.3 grams of magnesium with 9.45 ml of bromobenzene. you add the 4.83 grams of methyl benzoate to the freshly prepared grignard reagent to form an addition product. finally, after hydrolyzing the grignard addition product, you obtain 5 grams of the final product, triphenyl carbinol. what is the percent yield of triphenyl carbinol?

95.6 %

Explanation:

For this question, we will have 2 reactions, the formation of the grignard reagent and the formation of the alcohol. The first step then is the calculation of the maximum amount of the grignard reagent. For this, we have to convert the grams to moles and check the smallest value. To do this we have to take into account the following conversion ratios:

Molar mass of Mg = 24 g/mol

Molar mass of phenylmagnesium bromide ()= 157 g/mol

Density of bromobenzene= 1.5 g/mL

Molar ratio between Mg and  = 1:1

The smallest value is the mol of bromobenzene therefore 0.0402 mol of phenylmagnesium bromide would be produced.

The next step is repite the same steps for the reaction of formation of the alcohol. Therefore we have to find the moles of methyl benzoate, so:

Molar mass of methyl benzoate: 136.14 g/mol

The we have to divide by the coefficient of each reactive in the balance reaction. So:

Therefore the limiting reagent would be the phenylmagnesium bromide. Now, the molar ratio between the phenylmagnesium bromide and triphenyl carbinol is 2:1, so the amount of alcohol produced is 0.0201 mol triphenyl carbinol. The next step is the conversion from mol to grams of triphenyl carbinol:

Molar mass of triphenyl carbinol= 260.33 g/mol

Finally, we have to divide the obtanied solid by the calculated one:

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

c-14 is carbon-14, or radiocarbon.   it is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons.