Question

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

a) how many grams of aluminum sulfate would be formed if 250g h2SO4 completely reacted with aluminum ​

Answer 1

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

Step-by-step 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 =
`(mass)/(atomic mass of one mole)`

number of moles =
`(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.

`(1)/(3)`=
`(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.