Question:
The amount of calcium carbonate that reacts with 500 cc of 0.5 N hydrochloric acid is
The amount of calcium carbonate that reacts with 500 cc of 0.5 N hydrochloric acid is
Updated On: Jun 23, 2023
- 50 g
- 25 g
- 12.5 g
- 100 g
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The Correct Option is B
Solution and Explanation
Mass of HCl in 500 cc of 0.5 N HCI
$w = \frac{0.5 \times 36.5 \times 500}{1000} \left[ \because \:\:\: N = \frac{w \times 1000}{e wt.x V (in cc)}\right]$
$ = 9.125 \, g$
The balanced equation for the reaction is
$ { $\underset{\text{100 g}}{ {Ca}}C0_3$ + $\underset{\text{2 $\times$ 36.5 g}}{{2HCl}}$ -> CaCl_2 + H_2O + CO_2}$
${2 \times 36.5\, g}$ of HCI react with 100 g of $CaCO_3$
$ \therefore$ 9.125 g of HCI will react with
$ \frac{100 \times 9.125}{2 \times 36.5} = 12.5 \, g$ to $CaCO_3$
$w = \frac{0.5 \times 36.5 \times 500}{1000} \left[ \because \:\:\: N = \frac{w \times 1000}{e wt.x V (in cc)}\right]$
$ = 9.125 \, g$
The balanced equation for the reaction is
$ { $\underset{\text{100 g}}{ {Ca}}C0_3$ + $\underset{\text{2 $\times$ 36.5 g}}{{2HCl}}$ -> CaCl_2 + H_2O + CO_2}$
${2 \times 36.5\, g}$ of HCI react with 100 g of $CaCO_3$
$ \therefore$ 9.125 g of HCI will react with
$ \frac{100 \times 9.125}{2 \times 36.5} = 12.5 \, g$ to $CaCO_3$
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