Posted by Fai on Wednesday, December 25, 2013 at 6:44pm.
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%.
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