Question

can any one help me out I reward you all. thank in advance

When 135 grams of titanium tetrachloride react with an excess of water, as shown in the unbalanced chemical equation below, how many grams of hydrochloric acid will be produced? Please show all your work for the calculations for full credit.

TiCl4 + H2O yields TiO2 + HCl


this is my balance equation

TiCl4 + 2H2O ----> TiO2 + 4HCl

Answer 1

I got the answers I just not sure.

Answer 2

Good job balancing. :) Now we need to use stoichiometry to find the grams of HCl.

Answer 3

:)

Answer 4

We will need the molar mass of TiCl4 and HCl. Can you find those?

Answer 5

moles TiCl4 = 135 g /186.692 g/mol=0.712

moles HCl = 0.712 x 4 =2.85

Answer 6

Sweet, you are almost done. :)

Answer 7

135 g TiCl4 * (1 mol TiCl4 / 189.67 g TiCl4) * (4 mol HCl / 1 mol TiCl4) * (36.46 g HCl / 1 mol HCl) =
105g HC

Answer 8

O wait I for got one thing were the 189.67 comes from

Answer 9

I have one more question is different can you help me out pleaseeeeeee.

Answer 10

That looks fight to me. And I'll see what I can do for your other question.

Answer 11

Haha, *looks right

Answer 12

If 15.6 grams of iron (III) oxide reacts with 12.8 grams of carbon monoxide to produce 9.58 g of pure iron, what are the theoretical yield and percent yield of this reaction? Be sure to show the work that you did to solve this problem.

unbalanced equation: Fe2O3 + CO “yields”/ Fe + CO2



balance equation Fe2O3 +3 CO “yields” 2Fe + 3CO2

Answer 13

Ok, so it looks like to me that we need to first find the limiting reactant to determine the theoretical yield.

Answer 14

Do you know how to find which reactant is limiting?

Answer 15

nop

Answer 16

You will need to use stoichiometry to find the amount of iron produced using the amounts of products separately. Then the reactant producing the lesser amount of iron is limiting.

Answer 17

It is basically the reactant that will get used up first. And once you run out of one of the reactants, the reaction will stop.

Answer 18

Can you tell me how much theoretical iron can be produced from each of the reactants, assuming that it will all react?

Answer 19

If the CO is in excess, the iron oxide must be the limiting reagent

Answer 20

Awesome, iron (III) oxide is limiting. Now, the percent yield is the actual yield/ theoretical yield x 100%.

Answer 21

now what

Answer 22

You are given the actual yield in the problem and the theoretical yield is the amount produced by the limiting reactant.

Answer 23

Look i want to study this I am so confuse can you write that down in a way i could understand.

Answer 24

Ok, let's back up. How did you determine that CO is excess?

Answer 25

And do you understand that I am here because I want to help you, not confuse you?

Answer 26

so this will be the answers

Fe2O3 + 3CO ---> 2Fe + 3CO2

156g Fe3*(1mole Fe2O3/159.69g*3mole Co/1mole Fe203*28g Co=8.2g CO

128g CO*1mole CO/28g*1mol Fe2O3/3mole CO*159.68g FE2O3/1 mole=24.33g FE2O3.

so Fe2o3 is the limiting reactant

Answer 27

Ok, I think I see your confusion. You figured out the limiting reactant, but in a different way than what I suggested.

Answer 28

So, now you need to find how many grams of iron that can be produced from your limiting reactant.

Answer 29

This all that I got


Fe2O3 + 3CO ---> 2Fe + 3CO2

156g Fe3*(1mole Fe2O3/159.69g*3mole Co/1mole Fe203*28g Co=8.2g CO

128g CO*1mole CO/28g*1mol Fe2O3/3mole CO*159.68g FE2O3/1 mole=24.33g FE2O3.

so Fe2o3 is the limiting reactant

15.6 Fe2O3*1moleFe2O3/159.69*2mole Fe/1moleFe2O3*55.845Fe= 10.91g Fe

Percent yield= antual yield/Theoritical yield*100%

(9.58/10.91)*100%=87.8 This is the percent yield

Answer 30

Awesome. Good job.

Answer 31

than you so much god bless you!!! and a happy new year.

Answer 32

You're welcome. You too