3.364 g of hydrated barium chloride of bacl2.xh2o was dissolved in water and made up to a total volume of 250.0 ml. 10.00 ml of this solution required 46.92 ml of 2.530 x 10-2 m silver nitrate for complete reaction. calculate the value of x in the formula of hydrated barium chloride. given the net ionic equation for precipitation below.

Given:

Mass of hydrated barium chloride = 3.364 g

Total volume of barium chloride V(total)= 250 ml

Volume taken for titration V = 10 ml

Volume of AgNO3 consumed = 46.92 ml

Concentration of AgNO3 = 0.0253 M

To determine:

The value of x i.e. the water of hydration in BaCl2

Explanation:

The net ionic equation is-

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

Based on the reaction stoichiometry: Equal moles of Ag+ and Cl- combine to form AgCl

Moles of Ag+ consumed = moles of Cl- present

Moles of Ag+ = V(AgNO3) * M(AgNO3) = 0.04692 * 0.0253 = 0.00119moles

Moles of Cl- present = 0.00119 moles

Thus, 0.00119 moles of Cl- are present in 10 ml of the solution

Therefore, number of moles of Cl- in 250 ml would be-

= 0.00119 * 250 /10 = 0.02975 moles of cl-

Now:

2 moles of Cl- are present in 1 mole of BaCl2

Therefore, 0.02975 moles of Cl- correspond to- 0.02975 * 1/2 = 0.01488 moles of BaCl2

Molar mass of BaCl2 = 208.22 g/mol

Thus, mass of BaCl2 = 0.01488 moles * 208.22 g.mol-1 = 3.098 g

Mass of water of hydration = 3.364 - 3.098 = 0.266 g

# moles of water 'x' = .266/18 = 0.015 ≅ 1

Ans: Formula for hydrated barium chloride = BaCl2. 1H2O

a

explanation:

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer