Posted by Fai on .
1.64g of a mixture of cac03 and mgc03 was dissolved in 50ml of 0.8 m Hcl. The excess of acid required 16ml of 0.25< NAOH for neutralization. Calculate the percentage of cac03 and mgc03 in the sample
The right answer mgc03 52.02% and caco 47.98%
You need to use 3 solution of drug H containing respectively 50%, 20% and 5% to make 600ml of 10% drug H. How much of the 50% will you use?

math Steve helps me 
DrBob222,
I assume that 0.8m actually is 0.8M.
Typical equation is
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
HCl used initially = M x L = 0.8M x 0.05L = 0.04 mols to begin.
Back titrated = M x L = 0.25M x 0.016L = 0.004 mols excess HCl.
Amt HCl used = 0.04mols  0.004 mols = 0.036 mols. Convert that to mols MgCO3 and CaCO3 = 1/2 x 0.036 = 0.018 mols used for the 1.64 g of the mixture.
Let X = g CaCO3
and Y = g MgCO3

equation 1 is X + Y = 1.64g
equation 2 is
(X/molar mass CaCO3) + (Y/molar mass MgCO3) = 0.018

Solve those two equations simultaneously to find X and Y.
Then (X/1.64)*100 = %CaCO3 and
(Y/1.64)*100 = %MgCO3
#2
mL x % = mL x %
XmL x 50% = 600 mL x 10%.
Solve for X.
Note: You don't need any of the 20% and you don't need any of the 5%. 
math Steve helps me 
Steve,
From the previous solution, we ended up with
.776g CaCO3 = .00775 moles
.018.00775 = .0102 moles MgCO3
By mass, that's
.00775*100.09=.7756g CaCO3
.0102*84.31 = .8599g MgCO3
total mass: 1.6355g
47.4% CaCO3
That's pretty close, eh? Maybe roundoff errors make up the small difference.
Looks like you need to study how mixture problems work.
There are many solutions to the drug H problem, since you don't specify any other conditions other than the total volume and concentration.
x+y+z = 600
.05x + .20y + .50z = .10*600
Lots of solutions to that exist:
5% 20% 50%
440 140 20
500 50 50
520 20 60