Question

If 50 g of CaCO3 sample on heating gives 22.4 g of CaO, then identify the correct statement(s). I: % purity of CaCO3 sample is 80% II: 8.96 L of CO₂ is obtained at STP III: The weight of the impurities present in the sample is 10 g​

Answer 1

Answer:

CaCO3

→

Δ

CaO+CO

2

Molar mass of CaCO_3CaCO

3

= 100g/mol

Molar mass of CaO = 56g/mol

Let the pure sample of CaCO_3CaCO

3

= y grams.

Mass of CaO produced after decomposition = 22.4 grams

If 100g of pure sample of CaCO_3CaCO

3

produces 56g of CaO, then "y" gram of pure sample of CaCO_3CaCO

3

produces 22.4 grams of CaO.

Calculating for y, we get

y=\frac{100\times 22.4}{56}y=

56

100×22.4

y = 40 grams

% purity of CaCO_3CaCO

3

can be calculated by:

\% purity=\frac{\text{mass of pure sample}}{\text{given impure mass}}\times 100%purity=

given impure mass

mass of pure sample

×100

Putting the values, we get

\%purity=\frac{40}{50}\times 100%purity=

50

40

×100

% purity of CaCO_3CaCO

3

in the sample = 80%

Explanation:

I hope you understood this