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The mass of silver chloride, AgCl, that can be produced from the reaction is 2.76 g

We'll begin by calculating the number of mole of silver nitrate, AgNO₃ and calcium chloride, CaCl₂ in the solution.

For AgNO₃:

Volume = 120.0 mL = 120 / 1000 = 0.12 L

Molarity = 0.16 M

Mole of AgNO₃ =?

Mole = Molarity x Volume

Mole of AgNO₃ = 0.16 × 0.12

Mole of AgNO₃ = 0.0192 mole

For CaCl₂:

Volume = 120.0 mL = 120 / 1000 = 0.12 L

Molarity = 0.22 M

Mole of CaCl₂ =?

Mole = Molarity x Volume

Mole of CaCl₂ = 0.22 × 0.12

Mole of CaCl₂ = 0.0264 mole

  • Next, we shall determine the limiting reactant

2AgNO₃ + CaCl₂ —> 2AlCl + Ca(NO₃)₂

From the balanced equation above,

2 moles of AgNO₃ reacted with 1 mole of CaCl₂.

Therefore,

0.0192 mole of AgNO₃ will react with = 0.0192 / 2 = 0.0096 mole of CaCl₂

From the calculation made above, we can see that only 0.0096 mole of CaCl₂ out of 0.0264 mole given, is needed to react completely with 0.0192 mole of AgNO₃.

Therefore, AgNO₃ is the limiting reactant and CaCl₂ is the excess reactant.

  • Next, we shall determine the number of mole of AgCl produced from the reaction using the limiting reactant.

2AgNO₃ + CaCl₂ —> 2AlCl + Ca(NO₃)₂

From the balanced equation above,

2 moles of AgNO₃ reacted to produce 2 moles of AlCl.

Therefore,

0.0192 mole of AgNO₃ will also react to produce = 0.0192 mole of AgCl

  • Finally, we shall determine the mass of AgCl produced from the reaction

Mole of AgCl = 0.0192 mole

Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol

Mass of AgCl =?

Mass = mole × molar mass

Mass of AgCl = 0.0192 × 143.5

Mass of AgCl = 2.76 g

Thus, the mass of AgCl produced from the reaction is 2.76 g

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