posted by Fai .
1.64g of a mixture of cac03 and mhco3 was dissolved in 50ml of 0.8M hcl. The excess of acid required 16ml of 0.25M naoh for neutralization. Calculate the percentage of cac03 and mgc03 in the sample
Who helps me
16ml of .25M NaOH = .004 moles NaOH
So, .004 moles HCl was needed to neutralize it
50ml of .8M HCl = .04 moles HCl
That means that .036 moles HCl reacted with the CO3 stuff
.018 moles of CO3 stuff reacted with the HCl.
CaCO3 has a mol wt of 100.09
MgCO3 has a mol wt of 84.31
x/100.09 + (1.64-x)84.31 = .018
x = .776g CaCO3
.776g CaCO3 = .00775 moles
.018-.00775 = .0102 moles MgCO3
You don't say whether you want the %ages in terms of moles or weight, so I'll leave that calculation to you.