Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3 (s) + 2 HCl (aq) →  CaCl2(aq) + CO2 (g) + H2O(l). What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

Asked by Abhisek | 1 year ago |  143

##### Solution :-

0.75 M of HCl

≡ 0.75 mol of HCl are present in 1 L of water

≡ [(0.75 mol) × (36.5 g mol–1 )] HCl is present in 1 L of water

≡ 27.375 g of HCl is present in 1 L of water

Thus, 1000 mL of solution contains 27.375 g of HCl

Therefore, amt of HCl present in 25 mL of solution

$$\dfrac{ 27.375 \; g }{ 1000 \; mL } \; \times \; 25 \; mL$$

= 0.6844 g

Given chemical reaction,

$$CaCO_{ 3 }\; (s) \; + \; 2 \; HCl\; (aq) \; \rightarrow \; CaCl_{ 2 }\;(aq) \; + \; CO_{ 2 }\; (g) \; + \; H_{ 2 }O\; (l)$$

2 mol of HCl (2 × 36.5 = 73 g) react with 1 mol of $$CaCO_{ 3 }(100 g)$$

Therefore, amt of $$CaCO_{ 3 }$$ that will react with 0.6844 g

$$\dfrac{ 100 }{ 73 } \; \times \; 0.6844 \; g$$

= 0.9375 g

Answered by Pragya Singh | 1 year ago

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