Under the right conditions aluminum will react with chlorine to produce aluminium chloride. 2 Al +3 Cl2 - 2 AlCl3
How many grams of aluminum chloride can be produced when 157.0g of aluminum react with excess chlorine?
CORRECT ANSWER IS 775.9 g AICI3 but what are the steps to get that 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

the answers would be

the first one and the second one

the answer is b. dilution