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Express of `CO_(2)` is passed through 50 mL of 0.5 M calcium hydroxide solution. After the completion of the reaction, the solution was evaporated to dryness. The solid calcium carbonated was completely neutralized with 0.1 N hydrochloric acid. The volume of hydrochloric acid required is (At mass of carbon = 40)A. `200 mL`B. `500 mL`C. `400 mL`D. `300 mL` |
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Answer» Correct Answer - B `Ca(OH)_(2)+CO_(2)toCaCO_(3)+H_(2)O` 50 mL of 0.5 mL `Ca(OH_(2)) = (0.5)/(1000)xx50`mol = 0.025 mol 1 mol of `Ca(OH)_(2)` produces `CaCO_(3)=1` mol `:.` 0.025 mol of `Ca(OH)_(2)` will produce `CaCO_(3) = 0.025` mole `CaCO_(3)+2HCl toCaCl_(2)+H_(2)O+CO_(2)` 1 mole of `CaCO_(3)` is neutralized by HCl = 2 mol =0.25 mol `CaCO_(3)` will be neutralized by HCl `=2xx0.025` mol = 0.05 mol 100 mL of 0.1 N HCl contain HCl = 1 eq. =0.1 mol or 0.1 mol of HCl is present in HCl sol = 1000 mL `:.` 0.05 mole of HCl is present in HCl sol `:. (1000)/(0.1)xx0.05 mL = 500 mL` |
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