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?

Could somebody please help walk me through this?

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?

Could somebody please help walk me through this?

anonymous · Answer

first find theoretical yield (mass of AlBr3 formed from 8.56g of HBr)
just use the steps from the previous question but in this case moles of AlBr3=1/3 moles of HBr
then find % yield using the formula % yield=experimental yield (given in question)/theoretical yield x 100

anonymous · Answer

wait can we start fresh and over from step 1?

anonymous · Answer

ok 
first step find Mr of limiting reagent (reagent not in excess in this case HBr)
step 2
find moles of HBr
step 3
moles of AlBr3=1/3 moles of HBr (from equation)
step 4
find mass of AlBr3
this is the theoretical yield
divide the yield given in the question by this yield and multiply by 100 to find %

chmvijay · Answer

still you didnt get it ????

anonymous · Answer

no I can figure it out when someone gives me the steps, but on my own I don't know where to start

chmvijay · Answer

ok then let me try it

chmvijay · Answer

since Al(OH)3 is in excess reactant we can say that limiting reagen will be HBr right

anonymous · Answer

ok seems only logical

chmvijay · Answer

yaaa now can you tell the molecular weight of HBr and AlBr3

anonymous · Answer

how do I find it?

anonymous · Answer

molar mass?

chmvijay · Answer

yaaa

chmvijay · Answer

no u find the molar mass

rane · Answer

lol vijay

chmvijay · Answer

why LOL

anonymous · Answer

molar mass of HBr is 80.91g
molar mass of AlBr3 is 266.69g

rane · Answer

once u said yes next minute no...bt the answer is same

chmvijay · Answer

ok now 80.91*3 gram of HBr produces =266.69 g ram of AlBr3 then
 8.56 gram of Hbr should produce  = 266.69*8.56   /80.91 =28.21 gram of AlBr3 you get

chmvijay · Answer

28.21 gram of AlBr3 =for 100 % 
but u r getting 8.91 gram =8.91*100  /28.21 = 31.58 %

chmvijay · Answer

LOL @rane sometime it happens :) nobody is perfect

rane · Answer

i know no body is perfect calm down

anonymous · Answer

yes I get it, so if 8.56 grams of HBr react with an excess of Al(OH)3 to form 8.91 grams of AlBr3, this reaction’s percent yield is 31.58%

anonymous · Answer

Thank you!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

anonymous · Answer

it seems easier now

chmvijay · Answer

you are  welcome :)