8. Magnesium carbonate (MgCO3) and calcium carbonate (CaCO3) both decompose when

Question

8. Magnesium carbonate (MgCO3) and calcium carbonate (CaCO3) both decompose when heated, forming CO2 gas and the corresponding oxides. MgCO3(s) CO2(g) + MgO(s) CaCO3(s) CO2(g) + CaO(s)

When a particular mixture of MgCO3 and CaCO3 is heated, it releases 47% of its mass as CO2, so that the oxide products have 53%of the mass of the original sample. What mass percentage of MgCO3 was present in the original mixture? (HINT: The fact that the sample loses 47% of its mass does not depend on the original sample size, only on the relative amounts of the two carbonates. So you can start with any size sample you’d like.)

im studying for a chem exam and can't seem to figure this practice problem out. please help thx

Explanation / Answer

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

CaCO3(s) CO2(g) + CaO(s)

Let the weight of magnesium carbonate and calcium carbonate is 100g

so weight of CO2 = 47 g

weight of CaO + MgO = 53 g

As per the stoichiometry

If we have x grams of MgCO3 then we will have 100-x g of CaCO3

So moles of MgCO3= x / molecular weight = x/ 84

It will give x/ 84 moles of CO2

so mass of CO2 = x X 44 / 84 grams of CO2 .........(1)

Moles of CaCO3 = 100-x / 100

It will give 100-x / 100 moles of CO2

So mass of CO2 = moles X molecular weight = (100-x ) X 44 / 100 grams .....(2)

Total mass of CO2 = 47grams = (1) + (2)

Mass of CO2 = 47 =   x X 44 / 84    + (100-x ) X 44 / 100

1.068 X 84 X 100 = 100x + 84 ( 100-x)

8971.2 = 100x + 8400 - 84x

571.2 = 16x

x = 35.7

So mass % of MgCO3 = mass of MgCO3 X 100/ total mass = 35.7%

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