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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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Answer» i. Mol of `HCl=(1000xx0.75)/1000=0.75` Mass of `HCl=0.75xx36.5 g=24.375 g` Wt. of `25 mL HCl=24.375/1000xx25=0.6844` ii. According to equation: `2 mol` pf `HCl` (i.e., `2xx36.5 g=73 g`) `HCl` react completely with `CaCO_(3)=1 mol =100 g`. `:. 0.6844 g HCl` will react completely with `CaCO_(3)` `=100/73xx0.6844 g=0.938 g` |
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