Question

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?

Answer 1

so would I divide?

Answer 2

somebody, anybody?

Answer 3

@phi

Answer 4

How many moles of Al(OH)3 do yo have ?
How many moles of water is created ?

Answer 5

can you write the balanced equation first

Answer 6

the balanced equation would be 3HBr + Al(OH)3 ----> 3H2O + AlBr3

Answer 7

that equation tells you 1 mole of Al(OH)3 makes 3 moles of water (in theory)

Answer 8

ok

Answer 9

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

Answer 10

moles are molar mass?

Answer 11

*39.1 g

Answer 12

molar mass is the mass of one mole

Answer 13

how do I find out how many moles of Al(OH)3 I have?

Answer 14

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

Answer 15

for Al(OH)3 you need to add up the atomic weight of Al and 3 O's and 3 H's

Answer 16

I got 2.172 moles

Answer 17

yes, now find the molar mass of Al(OH)3

Answer 18

78.0036 moles

Answer 19

I would use 78 g/mole

you have 61.3 grams. How many moles is that ?

Answer 20

so I divide 78 by 61.3?

Answer 21

\[ \frac{61.3 \text{ g}}{78 \frac{\text{ g}}{mole }}=\frac{61.3 \cancel{g}\text{ mole}}{78 \cancel{ g}} \]

Answer 22

wait so next I'm suppose to turn 61.3/78 into a percentage?

Answer 23

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

Answer 24

yeah we just did that

Answer 25

right?

Answer 26

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

Answer 27

0.78589 moles of Al(OH)3

Answer 28

now I multiply that by 3?

Answer 29

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)

Answer 30

2.35769 moles of water

Answer 31

that is "theoretical"

how many moles of H2O did we actually make?

Answer 32

2

Answer 33

? scroll back up. You already figured this out

Answer 34

2.172

Answer 35

now make the fraction: moles created/ moles theoretical
and change to a percent

Answer 36

2.172/2.35769

Answer 37

yes

Answer 38

percent yield of a reaction is that changed to a percent

Answer 39

0.9212

Answer 40

or 92%

Answer 41

ooooo that was cool

Answer 42

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%

Answer 43

ty!!!!!!!!!!!!!!!!!!!!!!!!!