How much Al must be used to react with 652 g of NH4ClO4?

Question

3Al(a) + 3NH4ClO4(s) = Al2O3(s) + AlCl3(s) + 3NO(g) + 6H2O (g)

Answer 1

3Al(a) + 3NH4ClO4(s) = Al2O3(s) + AlCl3(s) + 3NO(g) + 6H2O (g)  Balanced equation 

moles of NH4ClO4 present = 652 g x 1mole/117.49 g = 5.53 moles

 

From balanced equation, mole ratio Al(s) : NH4ClO4 is 3 : 3 or 1:1, so moles Al needed = moles NH4ClO4 present

 

Moles Al(s) needed = 5.53 moles (to 3 significant figures)

Mass Al(s) needed = 5.53 moles x 26.98 g/mole = 149 g (to 3 significant figures)