What mass of `CaCO_(3)` is required to react completely with 25 ml of `0.75 MHCI`?A. `0.94 g`B. `9.4 g`C. `0.094 g`D. `0.49 g`

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

1.	What mass of `CaCO_(3)` is required to react completely with 25 ml of `0.75 MHCI`?A. `0.94 g`B. `9.4 g`C. `0.094 g`D. `0.49 g`
Answer» Correct Answer - A `CaCO_(3) + 2HCl to CaCl_(2) + CO_(2) + H_(2)` 25 mL of 0.75 M HCl `= (25)/(1000)Lxx(0.75 mol L^(-1))` `= 0.01875 mol` Moles of `CaCO_(3)` required `= ("Moles of HCl")/(2)` `= (0.01875)/(2) = 9.375xx10^(-3) mol` Mass of `CaCO_(3)` required `= 9.375xx10^(-3) molxx100gmol^(-1)` `= 0.9375g = 0.94g`

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What mass of `CaCO_(3)` is required to react completely with 25 ml of `0.75 MHCI`?A. `0.94 g`B. `9.4 g`C. `0.094 g`D. `0.49 g`

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