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?
Can someone please do it and show me in the simplest and easiest way/steps possible?

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? 
Can someone please do it and show me in the simplest and easiest way/steps possible?

anonymous · Answer

@haseebgul17

anonymous · Answer

any chance you remember how to do these?

anonymous · Answer

first find theoretical yield ofAlBr3
molar mass of Al[OH]3=27+[16+1]3=78
molar mass of AlBr3=267
according to equation
78g of Al[OH]3 will give AlBr3=267g
1g of...............................=267/78=3.4g
8.56g of..........................=3.4*8.56=26.8g
this is theoretical yield
actual yield =8.91g
%yield=8.91/26.8*100=33%