Question

1.64 g of mixture of CaCO3 and MgCO3 was dissolved in 50 ml of 0.8M HCl. The excess of acid required 16ml of 0.25 M NaOH for neutralisation .Calculate the percentage of CaCO3 and MgCO3 in the sample.

Answer 1

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Answer 2

Answer:

Explanation:

equivalent of caco3+equivalent of mgco3 = 36

moles+moles=36/2000

moles+moles=0.018

w/100+1.64-w/84 = 0.018

164-16w = 151.2

16w=12.8

w=0.8 gm

weight of caco3 = 0.8 gm

weight of mgco3 = 0.84 gm

percentage composition of caco3 = 0.8/1.64*100 = 48.79

percentage composition of mgco3 = 51.21