Question

A sample of CaCO3 is 50% pure. On
heating 1.12L of CO2 (at STP) is
obtained. Residue left (assuming non -
volatile impurity) is

Answer 1

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Total mass of residue left = 7.8 g

$\fbox{\texttt{\pink{Given:}}}$

The purity of sample of $\bold{CaCO_3}$ = 50%

$\fbox{\texttt{\blue{To\:find:}}}$

Total mass of residue left after heating.

$\fbox{\texttt{\red{How\:to\:Find:}}}$

Decomposition (reaction) of $\bold{CaCO_3}$ :-

$\bold{CaCO_3----)CaO+CO_2}$

From this reaction, it is clear that 1 mole of $\bold{CaCO_3}$ gives 1 mole of CaO and 1 mole of $\bold{CO_2}$.

$\therefore$ we can say the moles of $\bold{CO_2}$ produced is equal to moles of $\bold{CaCO_3}$ consumed.

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Now, according to the question,

At STP 1 mole of $\bold{CO_2}$ = 22.4 L

Then, no. of moles of $\bold{CO_2}$ = $\bold{\frac{1.12}{22.4}=0.05}$

So, no. of moles of $\bold{CaCO_3}$ = 0.05

Molar mass of $\bold{CaCO_3}$ = 100 g/mole

We have 0.5 mole = 0.5 × 100 = 5 g $\bold{CaCO_3}$

Now, molar mass of CaO = 56 g/mole

Total Mass of CaO remaining = $\bold{\frac{5\times{}56}{100}=2.8\:g}$

$\therefore$ Total mass of residue left = 5.0 + 2.8 = 7.8 g

Answer 2

Explanation:

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