Question

Under the right conditions aluminum will react with chlorine to produce aluminum chloride.

2 Al + 3 Cl2 - 2 AlCl3
How many grams of aluminum are needed to react completely with 11.727 liters of chlorine?

Answer 1

`m_(Al)=9.42gAl`

Step-by-step explanation:

Hello there!

In this case, according to the given chemical reaction:

2 Al + 3 Cl2 --> 2 AlCl3

Whereas there is a 2:3 mole ratio of aluminum to chlorine; it will be possible for us to calculate the required grams of aluminum by using the equality 22.4 L = 1 mol, the aforementioned mole ratio and the atomic mass of aluminum (27.0 g/mol) to obtain:

`m_(Al)=11.727LCl_2*(1molCl_2)/(22.4LCl_2)*(2molAl)/(3molCl_2) *(27.0gAl)/(1molAl) \\\\m_(Al)=9.42gAl`

Regards!