Question

Attention Please!
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6.5g of an impure sample of limestone liberates 2.2g of $\sf CO_2$ on strong heating. The persentage purity of CaC$\sf O_3$ in the sample is

A) 85.2%

B) 76.9%

C) 72.5%

D) 92.5%
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Answer 1

AnSwer:

Ans to your question is 76.9%.

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GiVen:

The reaction according to the above question is :-

• CaCo3→CaO + Co2

Now, let us calculate the number of moles per each element.

• Calcium carbonate :

➠Given mass /molar mass

➠6.5g/100g/mole

➠0.065 moles.

• Carbon dioxide :

➠Given mass /molar mass

➠2.2g/44g/mole

➠0.05 moles.

Now, as we know that,

★1 mole of CaCO3 gives us 1 mole of CO2,

Thus, 0.065 moles of CaCO3 will give CO2 :

➔1/1 × 0.065

➔0.065 moles of CO2 itself.

Hence,

★Percentage of purity :-

➡Experimental product/ Theoretical product × 100

➡0.05/0.065 × 100

➡76.9%

Therefore, the percentage purity of CaCO3 is 76.9%.

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Hope it helps uh Dear ❤

Answer 2

Answer:

• Option (B)=> [76.9%]

Explanation:

Above picture contain explanation.

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