Question

how many grams of silver chloride are produced when 45 g of calcium chloride react with excess silver nitrate?​

Answer 1

Answer:

= 116 g AgCl

Explanation:

CaCl2 + 2AgNO3 = Ca(NO3)2 + 2AgCl

Mass of AgCl;

 1 Mole of CaCl2 = 111 g

Moles of CaCl2;

= 45 g CaCl2 (1 mol CaCl2/111 g CaCl2)

= 0.405 moles

Mole Ratio = 2: 1

Therefore; moles ,of AgCl

  = 0.405 moles × 2

  = 0.810 moles

Thus;

Mass of AgCl = 0.810 moles × 143.5 g AgCl

                      = 116 g AgCl