Question

when 12.8 g of Al reacts with 5.40 g of Cl2, how many grams of AlCl3 can form?

Answer 1

Al +   Cl2 ------>   AlCl3

balancing above equation

2  Al + 3 Cl2 ------> 2 AlCl3

2* 12.8 g 5.4 g

2* 12.8 / 27 = 3* 5.4 / 35.5

no of moles aluminium no of moles chlorie

=0.948moles    = 0.4563 moles

as the concentration of chlorine is found to be less therefore chlorine is the limiting reagent and hence

  0.4563 moles of product is formed .

as the molar mass of ALCL3 = 133.34 g /mol

no of moles * molar mass = given mass

given mass = 133.34 g /mol *   0.4563 moles = 60.84 g