Calcium Carbonate reacts with aqueous HCl to give `CaCl_(2)andCO_(2)` according to the reaction,
`CaCO_(3)(s)+2HCl(aq)toCaCl_(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?

Explanation

Given chemical reaction
`CaCO_(3(s))+2HCl_((aq))to CaCl_(2(aq))+CO_(2(g))+H_(2)O_((f))`
Molarity= `("Wt")/("GM Wt")xx(1000)/(V(m))`
0.75= `("Wt")/(36.5)xx(1000)/(25)`
M=0.75 M
W=?
Wt= `(0.75xx36.5xx25)/(1000)`
= 0.6844 gms of HCl
From the equation
`1 " mole of" CaCO_(3) to 2 " moles of HCl"`
100 gms of `CaCO_(3) to 2xx36.5 ` gms of HCl
? `to 0.6844` gms of HCl
`:. " Mass of" " "CaCO_(3)=(0.6844)/(2x36.5)xx 100=0.9375 ` gms
GMwt=36.5
V=25 ml