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  • Calcium carbonate reacts with aqueous HCl to give CaCl_{2} and CO_{2} according to the reaction given below: CaCO_{3} (s) + 2HCl (aq) \rightarrow CaCl_{2}(aq) + CO_{2}(g) + H_{2}O (l)

Calcium carbonate reacts with aqueous HCl to give CaCl_{2} and CO_{2} according to the reaction given below:

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

What mass of  CaCl_{2} will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of CaCO_{3} ? Name the limiting reagent. Calculate the number of moles of CaCl_{2} formed in the reaction.

Answers (1)

Number of moles of HCI=250\; mL\times \frac{0.76\; M}{1000}=0.19\; mol

Given: Mass of CaCO_{3} = 1000 g

Number of moles of CaCO_{3} = \frac{1000\; g}{100\; g}=10\; \; mol

Referring to the equation, 1 mol of CaCO_{3} (s) needs 2 mol of HCI (aq), the ratio is 1:2.

 Hence, for 10 mol of CaCO_{3} (s), the number of moles of HCI required would be:

10\; Mol\; CaCO_{3}\times \frac{2\; Mol\; HCl}{1\; Mol\; CaCO_{3}}=20\; Mol\; HCl (aq)

But only 0.19 mol of HCI (aq) is available and therefore, HCI is limiting reagent.

Amount of CaCl_{2} will depend on the amount of HCl present. 2 mol of HCl (aq), gets 1 mol of CaCl_{2}, the ratio is 2:1.

Thus, 0.19 mol of HCl will get=0.19\; Mol\; HCl\times \frac{1\; mol\; CaCl_{2}}{2\; mol\; HCl}=0.095\; mol\; of\; CaCl_{2}

0.095 ×molar mass of  CaCl_{2}= 0.095 ×111 = 10.54 g

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