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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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Answer» nswr search What WOULD you like to ask? 11th Chemistry Some Basic Concepts of Chemistry Stoichiometry and Stoichiometric Calculations What is the maximum mass of... CHEMISTRY avatar 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 Share Study later 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. |
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