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CaCO(3) + 2HCI to CaCI(2) + CO(2)+ H(2)O...

`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 g

B

16.65 g

C

11.1 g

D

8.325 g

Text Solution

Verified by Experts

The correct Answer is:
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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