How many grams of BaCl2 is contained in 250 ml of solution if, after adding 40 ml of it to 25 ml of 0.1020 N.

How many grams of BaCl2 is contained in 250 ml of solution if, after adding 40 ml of it to 25 ml of 0.1020 N. AgNO3 solution for back titration consumed 15 ml of 0.09800 N. NH4CNS solution?

Given:
V solution (BaCl2) = 250 ml
V1 solution (BaCl2) = 25 ml
V solution (AgNO3) = 40 ml = 0.04 l
SN (AgNO3) = 0.1020 n
V solution (NH4SCN) = 15 ml = 0.015 l
SN (NH4SCN) = 0.09800 n

Find:
m (BaCl2) -?

Solution:
1) BaCl2 + 2AgNO3 => 2AgCl ↓ + Ba (NO3) 2;
AgNO3 + NH4SCN => AgSCN ↓ + NH4NO3;
2) n (NH4SCN) = CH (NH4SCN) * V solution (NH4SCN) / z = 0.098 * 0.015 / 1 = 0.00147 mol;
3) n rest. (AgNO3) = n (NH4SCN) = 0.00147 mol;
4) n total (AgNO3) = CH (AgNO3) * V solution (AgNO3) / z = 0.102 * 0.04 / 1 = 0.00408 mol;
5) n react. (AgNO3) = n total (AgNO3) – n rest. (AgNO3) = 0.00408 – 0.00147 = 0.00261 mol;
6) n1 (BaCl2) = n reag. (AgNO3) / 2 = 0.00261 / 2 = 0.001305 mol;
7) m1 (BaCl2) = n1 (BaCl2) * M (BaCl2) = 0.001305 * 208 = 0.27144 g;
8) 25 ml – 0.27144 g;
250 ml – x g;
m (BaCl2) = x = 250 * 0.27144 / 25 = 2.7144 g.

Answer: The mass of the BaCl2 solution is 2.7144 g.