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Question

What is the maximum mass of Al(OH)3 that can be prepared by reaction of 13.4 grams of AlCl3 with 10 grams of NaOH according to the following equation?
AlCl3 + 3NaOHAl(OH)3 + 3NaCl

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Solution

Moles of AlCl3=13.4133.5=0.1 moles
Moles of NaOH=1040=0.25 moles
AlCl3+3NaOHAl(OH)3+3NaCl
0.1 0.25
0.0167 0.0833
Since, NaOH is limiting 0.0833 moles of Al(OH)3 are formed.
Mass =0.0833×78=6.5 gm

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