Question

2Al+6HBr -> 2AlBr3+3H2. When 3.22 moles of Al reacts with 4.96 moles of HBr, how many moles of H2 are formed?

Answer 1

2 Al+6 HBr =  2 AlBr₃ + 3 H₂

2 moles Al ---------  6 moles HBr ----------- 3 moles H₂
3.22 moles Al ------ 4.96 moles HBr ----- ( moles H₂ )

moles H₂ = 4.96 x 3 / 6

moles H₂ = 14.88 / 6

= 2.48 moles of H₂

hope this helps!

Answer 2

2Al +6HBr ---> 3H2 + 2AlBr3 
determine which is limiting... HBr is limiting 
4.96 moles HBr ---> 1/2 (4.96) moles H2 
2.92 moles H2 gas