A20.7167 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

The mass of impure is 20.7167 g. The decomposition reaction is as follows:

Here, 1 mole of gives 1 mole of MgO.

Molar mass of  and MgO is 84.31 g/mol and 40.3044 g/mol respectively.

Converting number of moles in terms of mass,

Here, M is molar mass.

Since,

Thus,

On putting the values,

Rearranging,

Or,

...... (1)

Let the mass of impurity be  and mass of impure is 20.7167 g thus,

...... (2)

Also, mass of impure MgO is 16.8817 g thus,

...... (3)

On comparing equations (2) and (3),

Putting the value of in equation (1),

Or,

Or,

Or,

Or,

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

mention the spill to the teacher clean the area and throw the broken pieces away

explanation: