Question

50
g of magnesium carbonate (MgCO3)
sample decomposes on heating to give
magnesium oxide (MgO) and 11 g of
carbon dioxide (CO2). Percentage purity
of MgCO3 in the sample is
(A) 42%
(B) 25%
(C) 30%
(D) 58%​

Answer 1

ANSWER: HOPE IT HELPS YOU

EXPLANATION

Answer 2

Answer:

In the given sample of MgCO₃, the percentage purity of magnesium carbonate is 42%.

Therefore, the option (A) is correct.

Explanation:

The decomposition reaction of magnesium carbonate :-

MgCO₃  $\longrightarrow$ MgO + CO₂

The molar mass of magnesium carbonate = 84g/mol

The molar mass of magnesium oxide = 40g/mol

Molar mass of carbon dioxide = 44g/mol

44g/mol carbon dioxide is given by MgCO₃ = 84g/mol

11g carbon dioxide is given by MgCO₃ $=\frac{84}{44}\times 11$ = 21g

The weight of given sample of  magnesium carbonate = 50g

Percentage purity of a substance in a sample is calculated by dividing the mass of the pure chemical by the total mass of the sample, multiplied by 100.

Percentage purity of magnesium carbonate $=\frac{21}{50}\times 100 = 42\%$

Therefore, the percentage purity of magnesium carbonate is 42%.

To learn more about "Percentage purity"

https://brainly.in/question/34528

To learn more about "Percentage of substance in sample"

https://brainly.in/question/23771816