Question

A 20.7167 g sample of impure magnesium carbonate was heated to complete decomposition according to the equation mgco3(s) → mgo(s) + co2(g). after the reaction was complete, the solid residue (consisting of mgo and the original impurities) had a mass of 16.8817 g. assuming that only the magnesium carbonate had decomposed, how much magnesium carbonate was present in the original sample? answer in units of g. 013 5.0 points the reaction of 8.9 grams of fluo

Answer 1

The mass of impure $MgCO_{3}$ is 20.7167 g. The decomposition reaction is as follows:

$MgCO_{3}(s)\rightarrow MgO(s)+CO_{2}(g)$

Here, 1 mole of $MgCO_{3}$ gives 1 mole of MgO.

Molar mass of $MgCO_{3}$  and MgO is 84.31 g/mol and 40.3044 g/mol respectively.

Converting number of moles in terms of mass,

$n=\frac{m}{M}$

Here, M is molar mass.

Since, $n_{MgCO_{3}}=n_{MgO}$

Thus, $\frac{m_{MgCO_{3}}}{M_{MgCO_{3}}}=\frac{m_{MgO}}{M_{MgO}}$

On putting the values,

$\frac{m_{MgCO_{3}}}{84.31}=\frac{m_{MgO}}{40.3044}$

Rearranging,

$m_{MgO}=\frac{m_{MgCO_{3}}}{84.31}\times 40.3044$

Or,

$m_{MgCO_{3}}=\frac{m_{MgO}}{0.4780}$...... (1)

Let the mass of impurity be $M_{I}$  and mass of impure $MgCO_{3}$ is 20.7167 g thus,

$M_{MgCO_{3}}=(20.7167-M_{I})g$...... (2)

Also, mass of impure MgO is 16.8817 g thus,

$M_{MgO}=(16.8817-M_{I})g$...... (3)

On comparing equations (2) and (3),

$M_{MgO}=M_{MgCO_{3}}-3.835$

Putting the value of $M_{MgO}$ in equation (1),

$m_{MgCO_{3}}=\frac{M_{MgCO_{3}}-3.835}{0.4780}$

Or,

$0.4780 M_{MgCO_{3}}=M_{MgCO_{3}}-3.835$

Or,

$M_{MgCO_{3}}-0.4780M_{MgCO_{3}}=3.835$

Or,

$0.522 M_{MgCO_{3}}=3.835$

Or,

$M_{MgCO_{3}}=\frac{3.835}{0.522}=7.35 g$

Thus, magnesium carbonate present in the original sample is 7.35 g.