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Question

A mixture of calcium and magnesium carbonates weighing 1.4 g was strongly heated until no further loss of weight was perceived. The residue weighed 0.76 g. What percentage of MgCO3 was present in the mixture?

A
20.45
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B
18.55
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C
16.33
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D
22.4
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Solution

The correct option is A 20.45
CaCO3+MgCO3CaO+MgO
let mass of CaCO3 = x g
and mass of MgCO3 = 1.4x g

moles of CaCO3 = x100 = moles CaO produced


moles of MgCO3 = x84 = moles MgO produced

mass of CaO produced = moles * no. of moles = (x)56100

mass of MgO produced = moles * no. of moles = (1.4x)40100

so, total weight of product = (1.4x)56100 + (1.4x)40100 = 0.76
so, value of x = 1.356 g
percentage of CaCO3 = 1.3561.4100 = 96.8 %
percentage of MgCO3 = 3.14 %

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