A mixture of 1 mol of Al and 3 mol of Cl2 are allowed to react as 2AI(s)+ 3CI2(g)-2AICI3
a. How many moles of AICI3 are formed?
b. Identify the limiting reagent.
c. How many moles of excess reactant left unreacted?
a. According to equation, 2 mole Al reacts with 3 mole Cl2 to give 2 mole AICI3.
No. of moles of Al taken = 1 mol
.’. No. of moles of Cl2 consumed= 1.5 mol
No. of moles of AICI3 formed = 1 mol
b. Aluminium is the limiting reagent.
c. No. of moles of Cl2 remain unreacted = 3 – 1.5 = 1.5 mol
