Question

If 10.5 g copper chloride react with 12.4 g aluminum what is the limiting reactant

Answer 1

Answer: copper chloride

Explanation:

`3CuCl_2+2Al\rightarrow 2AlCl_3+3Cu`

`Moles=\frac{\text{Given mass}}{\text{Molar mass}}`

`{\text {moles of copper chloride}=(10.5g)/(134g/mol)=0.08`

`{\text {moles of copper aluminium}=(12.4g)/(27g/mol)=0.46`

From the balanced equation. it can be seen that 3 moles of copper chloride reacts with 2 moles of Al.

Thus 0.08 moles of copper chloride reacts with=
`\frac2}{3}* 0.08=0.05` moles of Al.

Thus copper chloride is the limiting reagent as it limits the formation of product and aluminium is the excess reagent as it is left unused. (0.46-0.05)=0.41 moles are in excess.

Answer 2

Molar mass of copper chloride is 134.45 g/mole, so the mole of 10.5 g copper chloride is 10.5/134.45 = 0.078 mole.  Molar mass of aluminum is 27 g/mole, so the mole of 12.4 g aluminium is 12.4/27 = 0.46 mole.  The formula of this reaction is as follows, 2Al + 3CuCl2 ⇒ 2AlCl3 + 3Cu. Thus the molar ratio of the reactants is Al:CuCl2 = 2:3. So to react with 0.078 mole of copper chloride, will need 0.078 x 2/3 = 0.052 moles of aluminum which is less than the given amount (0.46mole).  Therefore, copper chloride is the limiting reactant.