Question

1, Calcium carbonate (CaCO3) contains 40% calcium, 12% carbon and 48% oxygen by mass. Knowing that
the law of constant composition holds good, calculate the mass of the constituent elements present in 2 g
of calcium carbonate.
[Ans. Ca 0-8 g, C = 0-24 g.0096 gl
2. Apply the law of constant proportions to calculate the mass of oxygen that will be used up for combustion
of 5 g of H2 gas.
[Ans. 40 l
TONIC ATOMIC THEORY​

Answer 1

Answer:

1. According to law of definite proportions, in a chemical substance, the elements are always present in definite proportions by mass.

Given data:

Percentage of Ca present in CaCO3 – 40%

Percentage of C present in CaCO3  - 12%

Percentage of O present in CaCO3  - 48%

This means 100g of calcium carbonate contains 40g of Ca, 12g of C and 48g of O.

Similarly, 2g of Calcium carbonate contains 40% calcium , 12% carbon and 48 % oxygen by mass.

Let us find the masses of Ca, C and O present in 2g of CaCO3.

Mass of Calcium present in 2g of CaCO3 = 40/100 X2 = 0.8g

Mass of Carbon present in 2g of CaCO3 = 12/100 X 2 = 0.24g

Mass of Oxygen present in 2g of CaCO3 = 48/12 X 2 = 0.96g.

Explanation:

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