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Albert Harpreet Banerjee
Asked: 2022-11-08T14:16:00+05:30 In:

Calcium carbonate reacts with aqueous `HCl` to give `CaCl_(2)` and `CO_(2)` according to the reaction:
`CaCO_(3)(s)+2HCl(aq) rarr 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`?

Calcium carbonate reacts with aqueous `HCl` to give `CaCl_(2)` and `CO_(2)` according to the reaction:
`CaCO_(3)(s)+2HCl(aq) rarr 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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  1. Correct Answer – `0.94 g Ca Co_(3)`
    0.75 M of HCl `-=` 0.75 mol of HCl are present in 1 L water
    `-=[(0.75 mol)xx(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 sontains 27.375 g of HCl.
    `therefore` Amount of HCl present in 25 mL of solution
    `=(27.375 g)/(1000 mL)xx25 ml`
    =0.6844 g
    From the given chemical equation ,
    `CaCO_(3(s))+2HCl_(aq) to CaCl_(2(aq))+CO_(2(g))+H_(2)O_((l))`
    2 mol of HCl `(2xx36.5 =71 g)` react with 1 mol of`CaCO_(3)` (100 g)
    `therefore`Amount of `CaCo_(3)` that will react with 0.6844 g` (100)/(71)xx0.6844g`
    =0.9639 g

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