Question

What is the minimum mass of Al(OH)3 that can be obtained by reaction of 13.4 gm of AlCl3 with 10 gm NaOH acc to following equation (at. mass of Al=27) AlCl3 + 3 NaOH -----> Al(OH)3 + 3NaCl

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Some Basic Concepts of Chemistry

Stoichiometry and Stoichiometric Calculations

What is the maximum mass of...

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Asked on December 26, 2019 by

Amruta Gosalia

What is the maximum mass of Al(OH)

3

that can be prepared by reaction of 13.4 grams of AlCl

3

with 10 grams of NaOH according to the following equation?

AlCl

3

+ 3NaOH→Al(OH)

3

+ 3NaCl

MEDIUM

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ANSWER

Moles of AlCl

3

=

133.5

13.4

=0.1 moles

Moles of NaOH=

40

10

=0.25 moles

AlCl

3

+3NaOH⟶Al(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

Hope this will help you.