Question

Percentage purity of 150g of sample which gives 5.6 litre at STP on complete decomposition is :
A.11.33%
B.16.67%
C.23.77%
D.33.33%

Answer 1

Hint: The decomposition reaction is the reaction in which the process of breaking of a chemical compound occurs by the action of heat, by the action of light, or by the action of electricity which tends to give two or more products. When the heat is provided during the decomposition reaction then that reaction is known as thermal decomposition reaction. This reaction gives two or more products.

Complete step-by-step answer:To calculate the number of moles $$CaC{O_3}$$we have divide the given volume by 22.4L. so the formula which we have to use is the following:
Number of moles=$$\dfrac{{5.6L}}{{22.4L}} = \frac{1}{4}$$moles
So now calculate the molar mass of calcium carbonate. The calcium mass is 40, carbon is 12 and oxygen is 16. But we have 3 oxygen here so it will be 3 times of 16 that is 64. So on addition we get,
Molar mass = 40+12+64=100$$gmo{l^{ - 1}}$$
Now let us find the moles in grams by multiplying the moles by its molar mass:
Weight of $$CaC{O_3}$$=$$\dfrac{1}{4} \times 100 = 25$$g
Now the mass given to us in question of calcium carbonate is 150g. the formula of percentage purity is the following:
Percentage purity= $$\dfrac{{mass CaC{O_3}\operatorname{int} hesample}}{{mass of the sample}} \times 100$$
On substituting the values we get:
Percentage purity=$$\dfrac{{25}}{{150}} \times 100 = 16.67\% $$

So the correct answer is option B.

Note:Calcium carbonate is an organic compound which is found in minerals and rocks such as calcite. It comprises the majority of the pearls and shells of the organisms of marine. It is amild basic salt and is used in mild neutralisation reactions. It is also used as the dietary supplement as a source for calcium.