Respuesta :
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.
Answer: copper chloride
Explanation:
[tex]3CuCl_2+2Al\rightarrow 2AlCl_3+3Cu[/tex]
[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
[tex]{\text {moles of copper chloride}=\frac{10.5g}{134g/mol}=0.08[/tex]
[tex]{\text {moles of copper aluminium}=\frac{12.4g}{27g/mol}=0.46[/tex]
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=[tex]\frac2}{3}\times 0.08=0.05[/tex] 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.
