Question

When solutions of silver nitrate and magnesium chloride are mixed, silver chloride precipitates out of solution according to the equation 2AgNO₃(aq)+MgCl2(aq)→2AgCl(s)+Mg(NO₃)₂(aq).

What mass of silver chloride can be produced from 1.96 L of a 0.233 M solution of silver nitrate? Express your answer with the appropriate units.

The reaction described in Part A required 3.46 L of magnesium chloride. What is the concentration of this magnesium chloride solution? Express your answer with the appropriate units.

Answer 1

2AgNO₃(aq) + MgCl₂(aq) → 2AgCl(s) + Mg(NO₃)₂(aq)

Here 2 moles of silver nitrate produces 2 moles of silver chloride.

1 moles of silver nitrate will produce 1 moles of silver chloride.

Now,

1.96 L of 0.233 M solution of silver nitrate.

Molar mass of silver nitrate = 169.87 g/mol.

1 M solution = 169.87 g in 1000 mL

0.233 M solution = (0.233 x 169.87) g in 1000 mL

= 39.58 g in 1000 mL

Now, 1000 mL contains 39.58 g

so, 1.96 L (= 1960 mL) contains (39.58 x 1960) / 1000 g

1960 mL contains 77.58 g

hence 77.58 g = 77.58 g / 169.87 g/mol = 0.46 moles

So, 0.46 moles of silver nitrate will produce 0.46 moles of silver chloride.

Now,

Molar mass of silver chloride =143.32 g/mol

1 mole of silver chloride =143.32 g

0.46 moles of silver chloride = (143.32 x 0.46) g = 65.93 g