Question

06 On thermal decomposition of 600 g of lime stone
(calcium carbonate) 44.8 L of Co, gas is produced
at STP. The percentage purity of lime stone is
(1) 50%
(2) 33.33%
(3) 75%
(4) 66.66%​

Answer 1

answer : option (2) 33.33%

explanation : mass of limestone = 600g

molar mass of limestone = 100g/mol.

so mole of limestone , n = 600/100=6

see reaction...

$CaCO_3(\textbf{limestone})\rightarrow CaO+CO_2$

one mole of CaCO3 produces one mole of CO2 gas.

so, 6 moles of CaCO3 produce 6 moles of CO2 gas.

volume of 6 mole of CO2 gas , V = 6 × 22.4L = 134.4L

given volume of CO2 gas , v = 44.8L

% percentage purity of limestone = 44.8L/134.4L × 100

= 200/6

= 33.33%

hence, percentage purity of limestone is 33.33%