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80ml of hcl is added to 2.5gm of pure cac03 when the reaction is over, then 0.5gm cac03 is left. Find the normality of the acid

The answer 0.5N

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mL x N x m.e.w. = grams.
80 x N x 0.050 = 2.0

grams CaCO3. Started with 2.5, finished with 0.5, used 2.0 g
m.e.w. CaCO3 = 100/2000 = 0.050

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