Question

200g impure caco3 on heating gives 5.6l. Co2 gas at stp. Find the percentage of calcium in the limestone sample.

Answer 1

Hope you are satisfied with the answer

Answer 2

Answer:

Percentage of $calcium$ in the $CaCO_{3}$ sample is equal to 5.06%.

Explanation:

The chemical reaction of $limestone$ on heating:

            $CaCO_{3}$   →   $CaO+CO_{2}$

We know, Molar mass of  $CaCO_{3}$ $=100g$

$1mole$ of  $CO_{2}$ generated from $1mole$ of $CaCO_{3}$.

22.4L of $CO_{2}$ will be given by 100g of pure $CaCO_{3}$

5.6L of $CO_{2}$ is given by $=\frac{100*5.6}{22.4}=25.3g$ of pure $CaCO_{3}$

Therefore out of 200g of impure $CaCO_{3}$ only 25.3g of pure $CaCO_{3}$

Now if 100g of pure $CaCO_{3}$ has 40g of calcium

Then 25.3g of pure $CaCO_{3}$  has Ca $=\frac{40}{100}*25.3 =10.12g$

Percentage of calcium in $CaCO_{3}$ $=\frac{10.12}{200}*100$ $=5.06$%