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`CaCO_(3)` +` 2HCI to CaCI_(2)` + `CO_(2)`+ `H_(2)O` what will be the amount of `CaCI_(2)` when 10 g` CaCO_(3)` mL 0.75 M HCI is used in the reaction ?A. 83.25 gB. 16.65 gC. 11.1 gD. 8.325 g |
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Answer» Correct Answer - D `CaCO_(3)+2HCI to CACI_(2)+CO_(2)+CO_(2)+H_(2)O` 10 g `(200xx0.25)/(1000) ` `0.1"mol"underset(("Imiting reagment"))(0.15"mol")` `n_(CaCI_(2))` formed `= (0.15)/(2)` `W_(CaCI_(2))=(0.15)/(2)xx111=8.325` g |
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