Aluminum reacts with bromine to produce aluminum bromide according to the following equation:
Al (s) + Br2 (l) → AlBr3 (s) [unbalanced]

you perform this reaction using an excess of aluminum and 55.7 grams of bromine. You end up collecting 60.0 grams of product aluminum bromide. What is your percent yield?

Question

Aluminum reacts with bromine to produce aluminum bromide according to the following equation:
    Al (s)  +      Br2 (l)  →     AlBr3 (s)	[unbalanced]

you perform this reaction using an excess of aluminum and 55.7 grams of bromine. You end up collecting 60.0 grams of product aluminum bromide. What is your percent yield?

jhonyy9 · Answer

first of all you need balance this equation

do you know it how ?

jhonyy9 · Answer

Al (s)  +      Br2 (l)  →     AlBr3 (s)

so on the left side there are 2 Br -s and in the right side there are 3 Br-s

how you make there equilibrium ?

mchea · Answer

Thank you for your response!

jhonyy9 · Answer

yw

but how make the equilibrum ?

jhonyy9 · Answer

$\color{#0cbb34}{\text{Originally Posted by}}$ @jhonyy9
by what number you need multiplie the 3 and by what number the 2 to get the same result ?
$\color{#0cbb34}{\text{End of Quote}}$