How many grams of aluminum are required to produce 8.70 moles of aluminum chloride

Question

This incredibly simple question has stumped me. I can't get the right answer.

Answer 1

A good place to start is the balanced chemical equation:

 

    2 Al + 3 Cl₂ →  2 Al Cl₃

 

  The key thing to notice is that according to this equation the molar ratio of Al to Al Cl₃ is 2:2 =  1:1

 

  Assuming the Cl₂ is in excess, this means that 8.70 moles of Al are required to make  8.70 moles Al Cl₃ .

 

   However,   from the periodic table, it is found that one mole of Al is 26.98 g of Al.  

 

  So  the required mass of Al is  8.70 x 26.98 =   234.7 g.  

  Rounding to 3 significant digits gives    235 g.