Question

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

Answer 1

Answer:

$m_{Al}=9.42gAl$

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:

$m_{Al}=11.727LCl_2*\frac{1molCl_2}{22.4LCl_2}*\frac{2molAl}{3molCl_2} *\frac{27.0gAl}{1molAl} \\\\m_{Al}=9.42gAl$

Regards!