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?
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%.
To find out how much of the 50% solution you need to use, you can set up a proportion based on the desired concentration of the final mixture.
Let x be the amount of the 50% solution needed.
The total volume of the final mixture is 600 ml, and you want it to have a concentration of 10%. This means that the amount of drug H in the final mixture should be 10% of 600 ml, or 60 ml.
Now, let's set up the proportion:
Amount of drug H in the 50% solution / Total volume of the final mixture = Desired concentration of drug H
We can express the above proportion in the following way:
(0.50 * x) / 600 ml = 0.10
To solve for x, you can cross-multiply and then divide:
0.50x = 600 ml * 0.10
0.50x = 60 ml
x = 60 ml / 0.50
x = 120 ml
Therefore, you need to use 120 ml of the 50% solution to make 600 ml of 10% drug H.
To solve this problem, let's assume x represents the amount of the 50% solution of drug H that you will use.
First, let's calculate the amount of drug H in each solution:
Amount of drug H in the 50% solution = 0.5x
Amount of drug H in the 20% solution = 0.2(600 - x)
Amount of drug H in the 5% solution = 0.05(600 - x)
Since you want to make a total of 600 ml of a 10% drug H solution, the amount of drug H in the final solution can be represented as:
0.1(600) = 0.5x + 0.2(600 - x) + 0.05(600 - x)
Simplifying the equation:
60 = 0.5x + 120 - 0.2x + 30 - 0.05x
Combining like terms:
60 = 0.25x + 150 - 0.05x
Rearranging the equation:
0.25x - 0.05x = 150 - 60
0.2x = 90
Dividing both sides of the equation by 0.2:
x = 90 / 0.2
x = 450
Therefore, you will need to use 450 ml of the 50% solution of drug H.
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