Asked by Fai
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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The answer 0.5N
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Answers
Answered by
DrBob222
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
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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