What mass of `CaCO_(3)` is required to react completely with 25 ml of `0.75 MHCI`?

Answer

C

Explanation

The balanced chemical equation is:
`CaCO_(3)(s) + 2HCl (aq) to CaCl_(2)(aq) + CO_(2)(g) + H_(2)O(l)`
Let us first calculate the moles of HCl in 25 mL of 0.75 M HCl.
1000 mL of 0.75 M HCl contain HCl = 0.75 mol
25 mL of 0.75 M HCl contain HCl `=0.75/1000 xx 25`
=0.01875 mol
2 mol of HCl completely react iwth 1 mol of `CaCO_(3)`
0.01875 mol of HCl will completely react with
`=1/2 xx 0.01875`
=0.009375 mol
Mass of 0.009375 mol of `CaCO_(3)= 0.009375 xx 100 = 0.9375`g