Question

2.67g of alumininium choloride was dissloved in water and an excess of silver nitrate solution was added to give a perecipitate of silver chloride AlCl3+3AgNo3---Al(NO3)3+3AgCl3 what mass os silver choride precipitate would be formed?

Answer 1

2.67g of aluminium chloride was dissolved in water and an excess of silver nitrate solution was added to give a precipitate of silver chloride AlCl3+3AgNo3---Al(NO3)3+3AgCl3 what mass os silver chloride precipitate would be formed?

AlCl3+3AgNo3---Al(NO3)3+3AgCl3

$$\begin{array}{l} \nu(AlCl_3) = \frac{2.67}{27 + (35.5 * 3)} = 0.202 \, \text{mol} \\ \nu(AgCl) = 0.202 * 3 = 0.606 \, \text{mol} \\ m(AgCl) = 0.606 * 143 = 86.658 \, g \\ \end{array}$$ $$m(AgCl) = 86.658 \, g$$