anonymous
  • anonymous
What is the percent yield of a reaction that produces 39.1 grams of water when 61.3 grams of Al(OH)3 react with an excess of HBr?
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
so would I divide?
anonymous
  • anonymous
somebody, anybody?
anonymous
  • anonymous
@phi

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More answers

phi
  • phi
How many moles of Al(OH)3 do yo have ? How many moles of water is created ?
chmvijay
  • chmvijay
can you write the balanced equation first
anonymous
  • anonymous
the balanced equation would be 3HBr + Al(OH)3 ----> 3H2O + AlBr3
phi
  • phi
that equation tells you 1 mole of Al(OH)3 makes 3 moles of water (in theory)
anonymous
  • anonymous
ok
phi
  • phi
you need to find how many moles of Al(OH)3 you have, multiply by 3, and that is the theoretical amount of water you should get. that will be the denominator of a fraction you need to change 31.1 grams of water into # of moles (amount actually created) that will be the numerator of the fraction
anonymous
  • anonymous
moles are molar mass?
phi
  • phi
*39.1 g
phi
  • phi
molar mass is the mass of one mole
anonymous
  • anonymous
how do I find out how many moles of Al(OH)3 I have?
phi
  • phi
let's do water first: O 16 g + 2 H (1 g each) = 18 g per mole you have 39.1 g divide 39.1 g by 18 g/mole to get the number of moles of water
phi
  • phi
for Al(OH)3 you need to add up the atomic weight of Al and 3 O's and 3 H's
anonymous
  • anonymous
I got 2.172 moles
phi
  • phi
yes, now find the molar mass of Al(OH)3
anonymous
  • anonymous
78.0036 moles
phi
  • phi
I would use 78 g/mole you have 61.3 grams. How many moles is that ?
anonymous
  • anonymous
so I divide 78 by 61.3?
phi
  • phi
\[ \frac{61.3 \text{ g}}{78 \frac{\text{ g}}{mole }}=\frac{61.3 \cancel{g}\text{ mole}}{78 \cancel{ g}} \]
anonymous
  • anonymous
wait so next I'm suppose to turn 61.3/78 into a percentage?
phi
  • phi
to figure out what to divide notice you have 61.3 grams . You want to cancel the grams (by dividing by grams) and multiply by moles so you want to do grams * moles/gram you have 71 grams /mole which is "upside down". this is a clue to divide
anonymous
  • anonymous
yeah we just did that
anonymous
  • anonymous
right?
phi
  • phi
61.3/78 gives you the number of moles of Al(OH)3 now multiply that by 3 to get the number of moles of water you should get
anonymous
  • anonymous
0.78589 moles of Al(OH)3
anonymous
  • anonymous
now I multiply that by 3?
phi
  • phi
yes, the balanced equation would be 3HBr + Al(OH)3 ----> 3H2O + AlBr3 that equation tells you 1 mole of Al(OH)3 makes 3 moles of water (in theory)
anonymous
  • anonymous
2.35769 moles of water
phi
  • phi
that is "theoretical" how many moles of H2O did we actually make?
anonymous
  • anonymous
2
phi
  • phi
? scroll back up. You already figured this out
anonymous
  • anonymous
2.172
phi
  • phi
now make the fraction: moles created/ moles theoretical and change to a percent
anonymous
  • anonymous
2.172/2.35769
phi
  • phi
yes
phi
  • phi
percent yield of a reaction is that changed to a percent
anonymous
  • anonymous
0.9212
phi
  • phi
or 92%
anonymous
  • anonymous
ooooo that was cool
anonymous
  • anonymous
So the percent yield of a reaction that produces 39.1 grams of water when 61.3 grams of Al(OH)3 react with an excess of HBr is 92%
anonymous
  • anonymous
ty!!!!!!!!!!!!!!!!!!!!!!!!!

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