Question

8g of CaCO3 containing silica is treated with dilutr HCl soln to give 0.88g of Co2 gas find percentage by mass in initial sample

Answer 1

25

Answer 2

The reaction between calcium carbonate ($\text{CaCO}_3$) and dilute hydrochloric acid ($\text{HCl}$) is given by:

$\text{CaCO}_3\text{(s)} + 2\text{HCl}\text{(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O}\text{(l)} + \text{CO}_2\text{(g)}$

Silica ($\text{SiO}_2$) is an impurity and does not react with dilute $\text{HCl}$.

The molar mass of $\text{CaCO}_3$ is $40.08 (\text{Ca}) + 12.01 (\text{C}) + 3 \times 16.00 (\text{O}) = 100.09 \text{ g/mol}$. We can approximate it as $100 \text{ g/mol}$ for simplicity.

The molar mass of $\text{CO}_2$ is $12.01 (\text{C}) + 2 \times 16.00 (\text{O}) = 44.01 \text{ g/mol}$. We can approximate it as $44 \text{ g/mol}$.

From the balanced chemical equation, 1 mole of $\text{CaCO}_3$ produces 1 mole of $\text{CO}_2$. Using molar masses, 100 g of $\text{CaCO}_3$ produces 44 g of $\text{CO}_2$.

We are given that 0.88 g of $\text{CO}_2$ is produced. We can use the stoichiometry to find the mass of $\text{CaCO}_3$ that reacted. Let $m_{\text{CaCO}_3}$ be the mass of $\text{CaCO}_3$ in the initial sample.

$\frac{\text{Mass of }\text{CaCO}_3}{\text{Molar mass of }\text{CaCO}_3} = \frac{\text{Mass of }\text{CO}_2}{\text{Molar mass of }\text{CO}_2}$

$\frac{m_{\text{CaCO}_3}}{100 \text{ g/mol}} = \frac{0.88 \text{ g}}{44 \text{ g/mol}}$

$m_{\text{CaCO}_3} = \left(\frac{0.88}{44}\right) \times 100 \text{ g}$

$m_{\text{CaCO}_3} = 0.02 \times 100 \text{ g}$

$m_{\text{CaCO}_3} = 2 \text{ g}$

So, the mass of pure $\text{CaCO}_3$ in the 8 g sample is 2 g.

The percentage by mass of $\text{CaCO}_3$ in the initial sample is calculated as:

$\text{Percentage of }\text{CaCO}_3 = \left(\frac{\text{Mass of }\text{CaCO}_3}{\text{Total mass of sample}}\right) \times 100$

$\text{Percentage of }\text{CaCO}_3 = \left(\frac{2 \text{ g}}{8 \text{ g}}\right) \times 100$

$\text{Percentage of }\text{CaCO}_3 = \left(\frac{1}{4}\right) \times 100$

$\text{Percentage of }\text{CaCO}_3 = 25\%$