Answer:
528g AlCl3
Explanation:
***Remember that you need all the terms except "g AlCl3" to cancel out
The atomic mass of Al is about 27g and the atomic mass of AlCl3 is about 132g
[tex]\frac{54g Al}{1} *\frac{1 molAl}{27gAl}* \frac{2 mol AlCl3}{1 mol Al}*\frac{132g AlCl3}{1 mol AlCl3}[/tex]
You need to take all the numbers and multiply/divide
528g AlCl3 should be your answer
