How many grams of aluminu?m will be produced if 12.0g of Ba reacts with 9.0g of Al2(SO4)3

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

How many grams of aluminu?m will be produced if 12.0g of Ba reacts with 9.0g of Al2(SO4)3

Chemistry
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

The reaction gives one type of precipitate, i.e. BaSO4 then, # mole of Ba = 12 / 137.3 = 0.0874 mole then # of mole of SO4 2- = 9/342.2 (molar mass of Al2(SO4)3) = 0.0263 mole since mole ratio of Ba: Al2(SO4)3 = 3:1, therefore, Ba is in excess. Thus, only 0.0263 mol Ba reacted to give BaSO4 since 3 mole of BaSO4 will be produced, so # mole of BaSO4 produced = 3 * 0.0263=0.0789 mole gram of BaSO4 = 0.0789 * 233.4 = 18.42g// correct me if I'm wrong ;)
\[3Ba + Al2(SO4)3 \rightarrow 3BaSO4 + 2Al ^{3+}\]
This is a problem of limiting reactants

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question