Question

Please Help :D

Calculate the maximum number of moles and grams of iodic acid (HIO3) that can form when 697 g iodine trichloride reacts with 118.2 g of water.
ICl3 + H2O ICl + HIO3 + HCl [unbalanced]

1. mol
2. g

3. What mass of the excess reactant remains?

Answer 1

@wolfe8

Answer 2

1) Balance the equation
2) Find number of moles of both known substances (iodine trichloride and water)

Do those for now.

Answer 3

the equation is

2ICL3 + 3H2O ---> ICl + HIO3 + 5HCl

Answer 4

and of ICL3 i do 697/232.35 = 2.999

and do I multiply this answer by 2 because there are 2 molecules of it?

Answer 5

I don't think you have the molar mass of ICl3 correct. It's 1 I and 3 Cl

Answer 6

ya I is 126.9 + (25.45 x3)
ha i found my mistake, i multiplied. ok so its um 3.429

Answer 7

do I multiply by 2 now?

Answer 8

Cl is 35.34 http://www.ptable.com/

Answer 9

ok well on my lab book periodic table it has Cl as 35.45

Answer 10

35.45 I mean. But either way, you calculated ICl3 wrong.

Answer 11

... i did 25 not 35 woooow ok but after i get the answer do i multiply it by 2?

Answer 12

its 233.25 btw

Answer 13

No you're looking for the number of moles that is mass over molar mass.

Answer 14

i knoww i am saying that ICl3 is 233.25

mass over molar mass is 697/233.25 which equals 2.988. but then there are 2 molecules because we balanced the equation.

Answer 15

We are not looking for the number of molecules or atoms. You need the number of moles of water too. And then,
3) Divide each number of moles with their coefficients in front of their formulas
4) The less one belongs to the limiting reagent.

Answer 16

then what is number 1 and 2 asking? its in mol and then in g...

and for H2O I got 6.56

so i have to divide 2.988 by 2? and 6.56 by 3?

Answer 17

I was giving you the steps to solving this. So after dividing which is lower?

Answer 18

ICl3 is lower. but thats the limiting reagent.

but how do I use this to find out the answer to the problem....

Answer 19

Since it is the limiting value, it will be the amount you have available for this reaction. So the number of moles of the products will depend on this. Use the original number of moles of ICl3 and use the mole ratio method again.

Answer 20

so use the number i got for ICl3. and the proccess we dd was the mole ratio method?

Answer 21

Yup.

Answer 22

so wouldnt it still be 2.988....... or do we also divide by 2?

Answer 23

No I said use the original moles. We divide by the coefficients just to see which one will run out first.

Answer 24

ahhh ok so the moles owuld be 2.988?

Answer 25

Yes

Answer 26

ok dokes. so what now mr. wolf?

Answer 27

well i got the 1. which is 1.49, but not the 2. which is grams. and the 3.