Question

3HBr + Al(OH)3 ----> 3H2O + AlBr3 If 8.56 grams of HBr react with an excess of Al(OH)3 to form 8.91 grams of AlBr3, what is this reaction’s percent yield? Haven't done this in 4ever can someone please do it and show me in the simplest steps possible? I will be extremely grateful

Answer 1

@jim_thompson5910

Answer 2

Step 1 would be to balance it, it's already balanced so moving on...

you're asked for the percent yield:

\[percent yield =\frac{ |experimental-theoretical| }{ theoretical }*100 \]
(equation 1)

you're given the experimental yield (8.91 grams of AlBr3), you need to find the theoretical yield, so convert the masses to moles (n):

\[moles(n)=\frac{ mass(m) }{ Molarmass(M) }\]

(you only need to do this for HBr since Al(OH)3 is in excess, meaning HBr is the limiting reactant - limits the amount of product you can make)

now use the stoichiometric coefficients and build a relationship:

--------------------------------------------------------------
For a general equation: aA +bB -> cC

where a,b,c (lowercase) are the coefficients and A,B,C are the species,

\[\frac{ n _{A} }{ a }=\frac{ n _{B} }{ b }=\frac{ n _{C}}{ c }\]
---------------------------------------------------------------

In this case, you have, 3HBr + Al(OH)3 ----> 3H2O + AlBr3, so:

\[\frac{ n _{HBr} }{ 3 }=\frac{ n _{AlBr _{3}} }{ 1 }\]

plug in the moles you found and solve for moles of AlBr3.

now use equation 1 with the experimental and theoretical yields to find the percent yield.