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Question

267 g of AlCl3 reacts with 120 g of NaOH to give aluminium hydroxide. The produced aluminium hydroxide reacts with 3 mol of H2SO4 to give aluminium sulphate. Calculate the moles of aluminium sulphate produced in the reaction.
AlCl3+3NaOHAl(OH)3+3NaCl
2Al(OH)3+3H2SO4Al2(SO4)3+6H2O

A
0.5 mol
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B
0.7 mol
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C
0.4 mol
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D
1.2 mol
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Solution

The correct option is A 0.5 mol
AlCl3+3NaOHAl(OH)3+3NaCl
2Al(OH)3+3H2SO4Al2(SO4)3+6H2O

Moles of AlCl3=given massmolar mass=267133.5=2 molMoles of NaOH=given massmolar mass=12040=3 molFor AlCl3=given molesStoichiometric coefficient=21=2For NaOH=given molesstoichiometric coefficient=33=1

Since the ratio is lower for NaOH, it will be the limiting reagent.
3 mol of NaOH produces 1 mol of Al(OH)3

In reaction (ii),
For Al(OH)3=given molesStoichiometric coefficient=12=0.5For H2SO4=given molesstoichiometric coefficient=33=1
So, Al(OH)3 will be the limiting reagent.
2 mol of Al(OH)3 produce 1 mol of Al2(SO4)3
1 mol will produce 0.5 mol of Al2(SO4)3

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