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

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**, Wednesday, December 25, 2013 at 8:22pm
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**, Wednesday, December 25, 2013 at 8:50pm
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

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