Ncert Exercises & Solutions

1.35 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?

NEETprep answer:

0.75 M of HCl ≡ 0.75 mol of HCl are 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.
$∴$ Amount of HCl present in 25 mL of solution
= $\frac{27.375 g}{1000 mL} x 25 mL$
= 0.6844 g
From the given chemical equation,
${CaCO}_{3 (s)} + {HCl}_{(aq)} \overset{}{\to} {CaCl}_{2 (aq)} + {CO}_{2 (g)} + H_{2} O_{(l)}$

71 g HCl react with of 100 g CaCO3 .
$∴$ Amount of CaCO3 that will react with 0.6844 g = $\frac{100}{71} 0.6844 g$
= 0.9639 g

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