posted 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
Who helps step by step
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