Answer to Question #238781 in General Chemistry for Simon

Question

Silver nitrate reacts with calcium chloride in solution to form a silver chloride precipitate.

1. An 11.8 g sample of silver nitrate is added to 127.6 mL of 0.500 mol/L solution of calcium chloride. Which of the two reactants is limiting?
2. What mass of silver chloride should be formed?

2AgNO3(s) + CaCl2(aq) → Ca(NO3)2(aq) + 2AgCl(s)

Answer 1

Moles of AgNO3 = 11.8/169.87 = 0.07mol

Moles of CaCl2 = 127.6 × 0.5 × 10^-3=

0.064

2 mole of AgNO3 = 1 mol of CaCl2

= 0.07×2 = 0.14mol

Silver nitrate is the limiting reactant and silver chloride precipitates.

2. Molar Mass of silver Chloride =143.32 g

= 143.32× 0.14 = 20.1g