3.364g of hydrated barium chloride BaCl2 xH20 was dissolved in water and made up to a total volume of 250ml . 10ml of this solution required 46.92ml of 2.530 x 10^-2 M silver nitrate for compelte reaction. Calculate the value of x in the formula of hydrated barium chloride, given the net ionic equation for precipitation below
5Fe + MnO4 + 8H ---> 5Fe + Mn + 4H2O
2007-10-22 18:49:10 · 2 answers · asked by techie08 2 in Science & Mathematics ➔ Chemistry
Sorry the equation is Cl + Ag ---> AgCl
2007-10-22 19:10:14 · update #1
x/.04692 = 2.53 x 10^-2 M AgNO3, so x = .00119 mol AgNo3
.00119 mol AgNO3 (1mol Cl/1molAg) = .00119 mol Cl
.00119 mol Cl / 10 ml = x/ 250mL, x = .02975 mol Cl
.02975 mol Cl * (1 mol BaCl2 / 2 mol Cl) = .014875 mol BaCl2
BaCl2 = 208.226g/mol
.014875 (208.226) = 3.1g BaCl2
3.364 - 3.1 = .264g H2O / 18 = .0146 mol H2O
.0146 mol H2O / .014875 mol BaCl2 = 1
So BaCl2 * 1H2O
2007-10-22 19:07:25 · answer #1 · answered by icie 3 · 0⤊ 0⤋
Any resemblence between your ionic eqtn and the problem is purely coincidental.
2007-10-23 02:01:31 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋