Question

For the following reaction, 56.1 grams of silver nitrate are allowed to react with 26.0 grams of copper(II) chloride. silver nitrate (aq) + copper(II) chloride (s) silver chloride (s) + copper(II) nitrate (aq) What is the maximum amount of silver chloride that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

The balanced chemical equation for reaction between silver nitrate and copper chloride to form silver chloride and copper nitrate is shown below,

2 AgNO3 (aq) CuC12 (s) Cu(NO3)2 (aq) 2 AgCl (s)

Let is first calculate the number of AgNO3 and CUCl2 is used in the reaction. The molecular weight of AgNO3 and CuCl2 are, 169.87 g/mol and 134.452 g /mol

The number of moles of AgNO3 used = Weight of AgNO3 in g / molecular weight in g/mol

= 56.1 g / 169.87 g/mol = 0.33 mol

The number of moles of CuCl2 used = Weight of CuCl2 in g / molecular weight in g/mol

= 26 g / 134.452 g/mol = 0.193 mol

From the balanced chemical equation it is clear that 1 mole of CuCl2 requires 2 moles of AgNO3.

The ideal molar ratio of CuCl2 and AgNO3 is, 1/2 = 0.5

Actual molar ration of CuCl2 and AgNO3 is, 0.193 / 0.33 = 0.58

As the actual molar ratio is higher than the ideal molar ratio, CuCl2 is used in excess and limiting reagent is AgNO3. Therefore, the yield of the reaction should be calculated from the moles of AgNO3 used in the reaction.

From the balanced chemical equation, it can be seen that 2 moles of AgNO3 give 2 moles of AgCl. Therefore, 0.33 moles of AgNO3 should form 0.33 moles of AgCl. As the molecular weight of AgCl is 143.32 g/mol, the amount of AgCl formed in grams can be calculated using following formula,

Mass of AgCl formed = number of moles of AgCl formed x molecular weight of AgCl in g/mol

= 0.33 x 143.32

= 47.19 g

Thus maximum amount of AgCl formed would be 47.19 g

The formula for limiting reagent is AgNO3

The excess of reagent remains = the amount of reagent used - amount of reagent consumed

Here, amount of CuCl2 consumed = 0.5 x number of moles of AgNO3

= 0.5 x 0.33

  = 0.165 moles

The excess of reagent remains = 0.193 - 0.165 = 0.028 moles

The excess amount of CuCl2 reagent left in g = excess number of moles of CuCl2 x molecular weight of CuCl2

= 0.028 x 134.35

= 3.76 g

Therefore, amount of excess reagent remains after reaction is complete is 3.76 g