Question

Solve this Stoichiometry problem

Answer 1

Answer : The balanced equation is;

2Al + 3 $Br_{2}$ ----> 2Al$Br_{3}$.

So now, to calculate how many grams of $Br_{2}$ will be needed to completely convert 15 g of Al$Br_{3}$,

In the reaction 2 moles of Al and 3 moles of $Br_{2}$ will be needed to produce 2 moles of Al$Br_{3}$.

So, (15.0 g Al) X (1 mole Al / 26.98 g Al) X (3 moles $Br_{2}$ / 2 moles Al) X (159.81 g $Br_{2}$ / 1 mole $Br_{2}$)

= 133 g $Br_{2}$

Hence, 133g of $Br_{2}$ will be needed to completely convert 15 g of Al into Al$Br_{3}$.