Answer to Question #240559 in General Chemistry for maria

Question

1) What mass of silver chloride can be produced from 1.48L of a 0.300 M solution of silver nitrate?

Answer: mass of AgCl= 63.6 g

2)The reaction described in Part A required 3.39 L of calcium chloride. What is the concentration of this calcium chloride solution?

Answer: ?

Answer 1

1.48L 0.3 M solution contains 1.48 L *0.3 mol/L = 0.444 mol AgNO3

Equation

AgNO3 + Cl- → AgCl + NO3-

1 mol AgNO3 produces 1 mol AgCl

0.444 mol AgCl will be produced

Molar mass AgCl = 107.87 + 35.45 = 143.32g/mol

Mass of 0.444 mol AgCl = 0.444 mol * 143.32 g/mol = 63.63 g AgCl produced.