(i) 2Al + 6HCl → 2AlCl 3 + 3H 2
(ii) AlCl 3 + 3NaOH → Al (OH) 3 + 3NaCl
(iii) Al (OH) 3 + NaOH → NaAlO 2 + 2H 2 O
Above series of reactions are carried out starting with 18 g of Al and 109.5 g of HCl in first step and further 100 g of NaOH is added for step (ii) and (iii). Find out limiting reagent in each step and calculate the maximum amount of NaAlO 2 that can be produced in step (iii). (Assume reactions are taken in sequence and also that each reaction goes to 100% completion)
L.R. in step (I) | L.R. in step (II) | L.R. in step (III) | Moles of NaAlO2 | |
(a) | AI | AlCl3 | Al(OH)3 | 0.66 |
(b) | AI | Na(OH) | Al(OH)3 | 0.5 |
(c) | AI | AlCl3 | NaOH | 0.5 |
(d) | HCL | AlCl3 | NaOH | 0.5 |
Text Solution
Verified by ExpertsC
Mole of Al = 
= 
Mole of HCl = 
= 3
Moles of NaOH = 
= 2.5
2 Al + 6 HCl 
2 AlCl 3 + 3H 2
Initial mole 2/3 3 0 0
final mole 0 1 2/3
AlCl 3 + 3NaOH 
Al(OH) 3 + 3NaCl
Initial mole 2/3 2.5 0 0
final mole 0 2.5 – 2/3 
3 2/3
= 0.5
Al(OH) 3 + NaOH 
NaAlO 2 + 2H 2 O
Initial mole 2/3 0.5 0 0
final mole 0 0.5
NaAlO 2 = 0.5 moles.
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems