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3 years ago

2Al + 3H2SO4->Al2(SO4)3+H2

Chemistry

1 answer:

beks73 [17]3 years ago

8 0

287.39 grams of Al2(SO4)3 is formed if 250g H2SO4 completely reacted with aluminum.

Explanation:

Balanced chemical equation:

2Al + 3H2SO4->Al2(SO4)3+H2

Data given:

atomic mass of Al2(SO4)3 = 342.14 gram/mole

mass of H2SO4 given = 250 gram

atomic mass of one mole of H2SO4 = 98.079 grams/mole

number of moles = $\frac{mass}{atomic mass of one mole}$

number of moles = $\frac{250}{98.079}$

number of moles = 2.54 moles

from the equation it can be seen as

3 moles of H2SO4 will react to give 1 mole Al2(SO4)3

2.54 moles of H2SO4 will react to give x moles of Al2(SO4)3.

$\frac{1}{3}$ = $\frac{x}{2.54}$

x = 0.84 moles of Al2(SO4)3 is formed.

to calculate mass of the Al2(SO4)3 = Number of moles x atomic weight of one mole

= 0.84 x 342.14

= 287.39 grams of Al2(SO4)3 is formed.

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