35.0 g of BaCl2 is added to 1.0 L of 0.65 M of AgNO3. The following reaction occurs: 2 AgNO3(aq)+BaCl2(aq)→2 AgCl(s)+Ba(NO3)2(aq)?
Calculate the mass of the precipitate, AgCl, that will be produced and the moles of the reactant in excess that will remain in the solution after the reaction is complete.
- Roger the MoleLv 71 month ago
2 AgNO3 + BaCl2 → 2 AgCl + Ba(NO3)2
(35.0 g BaCl2) / (208.233 g BaCl2/mol) = 0.16808 mol BaCl2
(1.0 L) x (0.65 mol AgNO3/L) = 0.65 mol AgNO3
0.16808 mole of BaCl2 would react completely with 0.16808 x (2/1) = 0.33616 mole of AgNO3, but there is more AgNO3 present than that, so AgNO3 is in excess and BaCl2 is the limiting reactant.
(0.16808 mol BaCl2) x (2 mol AgCl / 1 mol BaCl2) x (143.3212 g AgCl/mol) =
48.2 g AgCl produced
(0.65 mol AgNO3 initially) - (0.33616 mol AgNO3 reacted) =
0.31 mol AgNO3 left over