Calcium carbonate reacts with aqueous HCl to give CaCl 2 and CO 2 according to the reaction- CaCO 3 s -2HCl aq - CaCl 2 aq - CO 2 g - H 2 O l -What mass of CaCO 3 is required to react completely with 25 mL of 0-75 M HCl -

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Standard IX

Chemistry

Balancing Chemical Equations by Partial Equation Method

Calcium carbo...

Question

Calcium carbonate reacts with aqueous $H C l$ to give ${C a C l}_{2}$ and ${C O}_{2}$ according to the reaction,

${C a C O}_{3} (s) + 2 H C l (a q) \to {C a C l}_{2} (a q) + {C O}_{2} (g) + H_{2} O (l)$ .

What mass of ${C a C O}_{3}$ is required to react completely with 25 mL of 0.75 M $H C l$ ?

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Solution

The mass of HCl is $\frac{0.75 \times 36.5}{1000} \times 25 = 0.684 g$ .

1 mole of calcium carbonate reacts with 2 moles of HCl.

Hence, the mass of calcium carbonate that will react completely with 0.684 g of HCl is

$\frac{100}{2 \times 36.5} \times 0.684 = 0.937 g$ .

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

The mass of HCl is 0.75×36.51000×25=0.684g.1 mole of calcium carbonate reacts with 2 moles of HCl.Hence, the mass of calcium carbonate that will react completely with 0.684 g of HCl is 1002×36.5×0.684=0.937g.

Calcium carbonate reacts with aqueous $H C l$ to give ${C a C l}_{2}$ and ${C O}_{2}$ according to the reaction,

${C a C O}_{3} (s) + 2 H C l (a q) \to {C a C l}_{2} (a q) + {C O}_{2} (g) + H_{2} O (l)$ .

What mass of ${C a C O}_{3}$ is required to react completely with 25 mL of 0.75 M $H C l$ ?

The mass of HCl is $\frac{0.75 \times 36.5}{1000} \times 25 = 0.684 g$ .

1 mole of calcium carbonate reacts with 2 moles of HCl.

Hence, the mass of calcium carbonate that will react completely with 0.684 g of HCl is

$\frac{100}{2 \times 36.5} \times 0.684 = 0.937 g$ .