[Navi00]

It's real simple, but you have to start somewhere I guess, I just need someone to take me through this step by step once and I should have it down :

<i>Calculate the amounts requested if 1.34 mol H202 completely react according to the following equation.

2H20 ---> 2H20 + O2

Moles of Oxygen Formed ?

Moles of Water Formed ?</i>

Thanks =P

[mike]

First work out the mole ratios:

H₂O₂ : H₂O : O₂

2 : 2 : 1

or

1 : 1 : 1/2

so if you have 1.34 moles of H2O2:

1.34 : 1.34 : 0.67

[Navi00]

Hmmm... How exactly did you get those mole ratio?

I see two H atoms two O atoms in H202 but I only see two H atoms and 1 O atoms in H20 so wouldn't it be 1 1/2?

Thank you =P

[mike]

According to your reaction equation, 2 moles of hydrogen peroxide react to form 2 moles of water and 1 mole of oxygen.

I see two H atoms two O atoms in H202 but I only see two H atoms and 1 O atoms in H20 so wouldn't it be 1 1/2?

I don't understand this bit??

[Navi00]

Erhmm.. just explain how you impliment the mole ratio into the equation. Sorry, having a headache today ><

Thanks =p

[mike]

Your equation:

2H2O2 -----> 2H2O + O2

this is saying that for every 2 moles of peroxide that react you will make 2 moles of water and 1 mole of oxygen.

It could also be written as;

H2O2 --------> H2O + 1/2 O2

which would be saying that for every mole of peroxide that decomposes you create 1 mole of water and half a mole of oxygen.

But lets go back to you original equation:

2H2O2 -----> 2H2O + O2

the ratio of peroxide to water to oxygen is:

H2O2 : H2O : O2

   2    :  2    :  1

so if you have 1.34 moles of H2O2:

1.34 : 1.34 : 0.67

[Navi00]

Ahh okay so I just take 1.34 mol H202 x     1 mol 02
                                                          ----------     =  0.67 of O2
                                                          2 mol H202




and



1.34 mol H202 x  2mol H20
                         -----------   =   1.34 mol of H20
                          2mol H202


I appreshate the help mike =p

[Navi00]

Here is my next question if you don't mind helping me visulize it a bit better :

Calculate the amounts requested if 3.30 mol Fe203 completely react according to the following equation.

Fe203 + 2Al ---> 2Fe +Al203

a. Moles of Al needed

b. moles of iron formed

c. moles of aluminum oxide formed

(I'm not trying to get out of homework, just trying to get a better mindset on this type of problem =P) Thanks for the help again  

[mike]

Ok, so use the info I gave you earlier. See if you can write the mole ratio for this one yourself first

[Navi00]

Alright... lets see =p

3.30 mol Fe2O3

Fe2O3 +2Al ---> 2Fe + Al2O3
1 : 2  :   2    :    2  :   1 : 2

2Al = 27.0*2 = 54.0

54.0mol Al x 2 mol Al
                  -----------    =  108 mol of Al needed
                   1 mol Al203
                   

Hmmm.... is that right?

[mike]

No.

You are mixing moles and molecular weight and making the answer unnecessarily complicated.

Fe2O3 +2Al ---> 2Fe + Al2O3

Fe2O3 : Al : Fe : Al2O3

    1    : 2  : 2  :   1

so 1 mole of iron oxide reacts with 2 moles of aluminium to produce 2 moles of iron and 1 mole of aluminium oxide.

Therefore:

3.30 : 6.60 : 6.60 : 3.30

so 3.30 moles of iron oxide reacts with 6.60 moles of aluminium to produce 6.60 moles of iron and 3.30 mole of aluminium oxide.

Right?

ps don't try this reaction at home

[Navi00]

Hmm... it seems to be asking for the answers in grams now, but lets see :

Fe2O3 +2Al ----> 2Fe + Al2O3

    1        2           2        1




3.30 mol of Fe2O3 x    2mol Al
                                ----------        = 6.60 mol of Al Required
                                1 mol Al2O3


3.30 mol of Fe2O3    x   2 mol Fe
                                 ------------      = 6.60mol of Fe formed
                                  1 mol Fe2O3


3.30 mol Fe2O3    x   2 mol Al
                               -------------    = 6.60 mol of Al203 formed
                               1 mol Fe2O3

Is that right? \ =

[mike]

3.30 mol of Fe2O3 x    2mol Al
                                ----------        = 6.60 mol of Al Required
                                1 mol Al2O3


3.30 mol of Fe2O3    x  2 mol Fe
                                ------------      = 6.60mol of Fe formed
                                  1 mol Fe2O3


3.30 mol Fe2O3    x  2 mol Al
                              -------------    = 6.60 mol of Al203 formed
                              1 mol Fe2O3

Why are you setting it out like this? Is this how the book does it? I haven't seen this before. The ratio of iron oxide to aluminium oxide is 1 : 1, you have discovered this yourself already. So your calculation is wrong because you have calculated that 3.30 moles of iron oxide produces 6.60 moles of aluminium oxide, this is a ratio of 1:2 not 1:1!!

If the ratio of reactants is 1 : 2 (ie iron oxide : aluminium) then every 1 mole of iron oxide reacts with 2 moles of aluminium, that's it no calculation!

Fe2O3 :  Al :  Fe :    Al2O3

    1    :    2   :       2   :     1

or

3.30 : 6.60 :  6.60  :  3.30                (moles)

that's it, that is the answer you produce 6.60 moles of Fe and 3.30 moles of aluminium oxide.

[Navi00]

Hmmm... I guess my book is strange, it has them setup all werid.

Alright, than to convert those to grams you'd just do this? :

 
27.0g Al / 6.60 mol Al = 4.09g


&


102g Al203 / 3.30 Al203  31.0g

[mike]

n = m / M

where:

n = number of moles (mol)
m = mass (g)
M = molecular weight (g.mol-1)

Example:

Al

M = 27 g.mol-1
n = 6.60 mol

n = m/M

therefore:

m = nM = 6.60 x 27 = 178.2 g