A+B→A3B2 (unbalanced) A3B2+C→A3B2C2 (unbalanced) Above two reactions are carried out by taking 3 moles each of A and B and by taking one mole of C. Then which option(s) is/are correct?
12 mole of A3B2C2 is formed.
12 mole of A3B2 is left finally.
3A+2B→A3B2A/c to3 moles2 moles1 molestiochilometryGiven:3 moles3 moles−
3 moles of A react with 2 moles of B to give 1 mole of A3B2. Hence 1 mole of B would be left unreacted.
A3B2+2C→A3B2C21 mole2 mole1 mole(A/c to stoichiometry)1 mole1 mole−(Given)↓(from abovereaction)
2 moles of C reacts with 1 mole A3B2 to give 1 mole A3B2C2. So, 1 mole of C will react with 12 mole of A3B2 to give 12 mole of A3B2C2. So, 12 mole of A3B2 is left.