Question

Clay found in a lake at Bhopal contains 15% by mass moisture along with C a C O 3 and non volatile S i O 2 . This on heating loses a part of its moisture, but C a C O 3 is completely converted into CaO. The partially dried sample now contains 7.3% moisture and 51.5% S i O 2 . What is the mass percentage of C a C O 3 in the original sample of clay approximately?

Answer 1

Explanation:

Clay found in a lake at Bhopal contains 15% by mass moisture along with C a C O 3 and non volatile S i O 2 . This on heating loses a part of its moisture, but C a C O 3 is completely converted into CaO. The partially dried sample now contains 7.3% moisture and 51.5% S i O 2 . What is the mass percentage of C a C O 3 in the original sample of clay approximately?

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Answer 2

The mass% of CaCO₃ in the original sample of clay is 50%

GIVEN

Contents of sample - CaCO₃, SiO₂, Moisture.

Percentage weight of water in original sample = 15%

Percentage weight of water in the dried sample - 7.3%

Percentage weight of SiO₂ = 51.5%

TO FIND

Mass percentage of CaCO₃ in the original sample

SOLUTION

We can simply solve the above problem as follows -

Let the mass% Of CaCO₃ be, x

Let the total mass of clay = 100 grams

So,

Mass of water (moisture) = 100-15 = 85 grams

Mass of SiO₂ = 85-x grams

AFTER HEATING-

Mass% of water = 7.3%

Mass% of SiO₂ = 51.5%

Mass% of CaCO₃ = 100(51.5-7.3.) = 41.2%

Since Silicon dioxide is non-volatile, the mass will remain in both dried and orignal samples.

So,

$\frac{51.5}{100} \times weight \: of \: dried \: sample = 85 - x$

$weight \: of \: dried \: sample \: = \frac{(85 - x) \times 100} {51.5}$

So, the mass of CaO will be -

$= \frac{(85 - x) \times 100}{51.5} \times \frac{41.2}{100}$ (equation 1)

It is given that CaCO₃ is completely reduced to CaO

So, moles of CaO in the dried sample = Moles of CaCO₃ in the original sample

Moles= Given mass/molar mass

Given mass = Number of moles × molar mass

So, Mass of CaO = (x/100) × 56 .......(equation 2)

Equating equation 1 and equation 2 we get,

$\frac{(85 - x) \times 100}{51.5} \times \frac{41.2}{100} = \frac{x}{100} \times 56$

x = 50

Since we had taken the mass of original sample of clay is 100 grams

Mass% of CaCO₃ = 50%

Hence, the mass% of CaCO₃ in the original sample of clay is 50%

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