Answer:
Ok
The Limiting reagent is Aluminum. Its not in excess. So the reaction ends once its exhausted. So all Aluminum will be used up
From the question ....
157g of Al
Find the number if reacting Moles...
M=mass/molar mass
M= 157/27
=5.81moles of Aluminum is reacting.
Now looking at the equation of reaction(Always make sure its balanced)...
2moles of Al react to produce 2 Moles of AlCl3
Their Mole ratio is equal
Since their Mole ratio is equal 2:2.... That means... That 5.81moles of AlCl3 would be produced because the mole of Al is 5.81
So We now know the Moles of AlCL3 produced
getting the Mass won't be difficult.
Again
Mole=Mass/Mm
Mass = Mole x Mm(Molar Mass)
Mass of AlCl3 = 5.81 x (27 + 35.5x3)
Note: The arithmetic done in the parenthesis is the molar mass of AlCl3
Mass of AlCl3 = 5.81 x 133.5
=775.63g of AlCl3.
Have a great Day!!!!
