NEETChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
What mass of 95% pure CaCO3 will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction?
CaCO3(s) + 2HCl(aq) CaCl2(aq) + Co2(g) + 2H2O(l)
[Calculate upto second place of decimal point]
- A1.25 g
- B1.32 g
- C3.65 g
- D9.50 g
View written solutionFree
Correct answer: B
Let m gram mass of CaCO3 is required
Pure CaCO3 in m gram $$ = {{95} \over {100}} \times m$$
Moles of CaCO3 $$ = {{95} \over {100}} \times {m \over {100}}$$
Moles of HCl required = 2 $\times$ moles of CaCO3
$$ = 2 \times {{95} \over {100}} \times {m \over {100}}$$
$$2 \times {{95} \over {100}} \times {m \over {100}} = {{50} \over {1000}} \times 0.5$$
$$m = 1.315\,g \approx 1.32\,g$$
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