Magnesium carbonate (MgCO3) is used as a source of carbon dioxide gas for certai

Question

Magnesium carbonate (MgCO3) is used as a source of carbon dioxide gas for certain organic reactions where the system needs to be closed.

A. Give the balanced equation for the formatting of carbon dioxide and a second product that is a solid from magnesium carbonate.

B. The carbon dioxide that is formed is piped into a reaction vessel containing a Grignard reagent. The balanced equation is shown below. How much carbon dioxide would need to be generated to form 10 g of C6H5CO2H? C6H5MgCl + CO2 +HCL C6H5CO2 + MgCl2

C. To generate this many grams of carbon dioxide, how much magnesium carbonate would you need to start with?

D. If carbon dioxide and hydrochloric acid are used in excess, and you begin with 5 g of C6H5MgCl, what is the theoretical yield of C6H5CO2H? If you generate 2.5 g of C6H5CO2H what is the % yield?

Explanation / Answer

Magnesium carbonate (MgCO3) is used as a source of carbon dioxide gas for certain organic reactions where the system needs to be closed.

A. Give the balanced equation for the formatting of carbon dioxide and a second product that is a solid from magnesium carbonate.

Solution :-

Balanced reaction equation is as follows

MgCO3(s) --------- > MgO(s) + CO2(g)

B. The carbon dioxide that is formed is piped into a reaction vessel containing a Grignard reagent. The balanced equation is shown below. How much carbon dioxide would need to be generated to form 10 g of C6H5CO2H? C6H5MgCl + CO2 +HCL C6H5CO2 + MgCl2

Solution :-

Using the mole ratio of the C6H5CO2H and CO2 we can calculate the amount of the CO2 needed.

(10 g C6H5CO2H * 1 mol / 122.1213 g)*(1 mol CO2 / 1 mol C6H5CO2H) = 0.08189 mol CO2

Now lets convert moles of CO2 to its mass

Mass of CO2 = moles * molar mass

                      = 0.08189 mol * 44.01 g per mol

                      = 3.60 g CO2

So the mass of CO2 needed = 3.60 g

C. To generate this many grams of carbon dioxide, how much magnesium carbonate would you need to start with?

Solution :-

Using the mole ratio of the CO2 and MgCO3 we can find the mass of MgCO3 needed to give 3.60 g CO2

Mole ratio is 1 : 1

Therefore

Moles of MgCO3 needed are same as moles of CO2

Moles of MgCO3 needed = 0.08189 mol

Mass of MgCO3 = moles * molar mass

                            = 0.08189 mol * 84.3139 g per mol

                            = 6.90 g MgCO3

So the mass of MgCO3 needed = 6.90 g

D. If carbon dioxide and hydrochloric acid are used in excess, and you begin with 5 g of C6H5MgCl, what is the theoretical yield of C6H5CO2H? If you generate 2.5 g of C6H5CO2H what is the % yield?

Solution :-

Thereoretical yield is calculated using the mole ratio of the C6H5MgCl and C6H5CO2H

(5 g C6H5MgCl* 1 mol /136.8619 g )*(1 mol C6H5CO2H/1mol C6H5MgCl) = 0.03653 mol C6H5CO2H

Now lets find the mass of C6H5CO2H

Mass of C6H5CO2H = moles * molar mass

                                  =0.03653 mol * 122.1213 g per mol

                                  = 4.46 g C6H5CO2H

Therefore the theoretical yield of C6H5CO2H is 4.46 g

Now lets calculate the percent yield

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

             = (2.5 g / 4.46 g)*100%

             = 56.05 %

Therefore the percent yield is 56.05 %

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