`2Al_((s))+3H_2SO_(4(aq)) rarr Al_2(SO_4)_(3(aq))+3H_(2(g))`
`Al_2(SO_4)_3` is formed when we add `H_2SO_4` acid to Al solid.
`Al:Al_2(SO_4)_3 = 2:1`
Molar mass of Al = 27g/mol
Amount of Al reacted `= 270/27 = 10mol`
Amount of `Al_2(SO_4)_3` formed `= 1/2xx10 = 5mol`
Molar mass of `Al_2(SO_4)_3` = 342g/mol
Mass of `Al_2(SO_4)_3` obtained `= 342xx5 = 1710g`
So we can get a maximum of 1710g of `Al_2(SO_4)_3` .
- There is enough Sulphuric acid for the complete reaction
1710g of aluminium sulfate AI2(SO4)3
rams of aluminum sulfate , will be produced from the complete reaction of
xgAl2(SO4)3 / (54+ 96+192) = 270gAl / 2 x 27
- now we cross multiply
(54+96+192) = 342
342 x 270gAL= 92340
92340/ (2x27) = 1710
xgAl2(SO4)3 = 1710g
1710g grams of aluminum sulfate Al2(SO4)3 will be produced.