I made a typo so I've erased my entire response and posted the correct answer.
You know your answer can't be right because you have 4 g K2CO3 but the entire sample is only 1.2 g and 4 g is more than the you started with
My calculation 20cm3/100cm3 = 0.2
Yes, 0.2 is the factor and the mass of the sample is 0.2*1.2 = 0.24g or another way to look at it is 1.2 x (20/100) = 0.24g titrated. However, you only need the factor.
x/53 + (0.2g-X) + y/69 = 0.004
This titration is for 0.2 of the original or 0.2x and 0.2y; therefore,
(0.2x/53) + (0.2y/69) = 0.004
Also note that where you have substituted 0.2-x for x it should be 1.2-y. Since x + y = 1.2 then x = 1.2-y
0.2x/53 + 0.2y/69 = 0.004
3.773E-3*x + 2.898E-3*y = 0.004
Substitute for x = 1.2-y
3.77E-3(1.2-y) + 2.898E-3*y = 0.004
I will let you finish this but the answer is 0.6 g Na2CO3 and 0.6g K2CO3. By the way, I used 69.1 for K2CO3 that will not affect the answer much.
0.01886g-0.01886x+ 0.01449x =0.004g
x = 0.000228/0.00437
x= 0.052/1.2 x 100
x = 4g k2c03
1g - 4g = 6g na2c03 Who check for me
1.2g mixture of na2c03 and k2c03 was dissolved in water to form 100cm3 of solution, 20 cm3 of this solution required 40cm3 of 0.1 hcl for neutralization. Calculate the weight na2c03 and k2c03 in the mixture?
My calculation 20cm3/100cm3 = 0.2
x/53 + (0.2g-X) + y/69 = 0.004
0.01886g-0.01886x+ 0.01449x =0.004g
x = 0.000228/0.00437
x= 0.052/1.2 x 100
x = 4g k2c03
1g - 4g = 6g na2c03 Who check for me
1 answer