`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 ?

Options

A. 83.25 g
B. 16.65 g
C. 11.1 g
D. 8.325 g

Answer

D

Explanation

`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