Posted by **Zawng** on Tuesday, September 10, 2013 at 10:37am.

A 28.4378 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 17.6846 g. Assuming that only the magnesium carbonate had

decomposed, how much magnesium carbonate was present in the original sample?

Answer in units of g

- Chemistry -
**Jai**, Tuesday, September 10, 2013 at 1:21pm
I'll give you the steps on how to solve:

(1) Get the molecular weights (MW) of MgCO3 and MgO.

-> I think you know how to do this one, you just go to periodic table and find the masses of the individual atoms and add them according to the chemical formulas.

(2) Let x = grams of PURE MgCO3 in the 28.4378

-> Therefore, we can say that 28.4378 - x = total or original impurities

(3) Using x, solve for the mass of MgO produced. Follow the formula:

-> x (1 mol MgCO3 / MW MgCO3) * (1 mol MgO / 1 mol MgCO3) * (MW MgO / 1 mol MgO) = (MW MgO / MW MgCO3) * x

(4) Then, add the mass of original impurity with the mass of MgO produced, which is equal to 17.6846

-> (28.4378 - x) + ((MW MgO / MW MgCO3) * x) = 17.6846

(5) Finally, solve for x.

Hope this helps~ :3

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