Question

Stoichiometry help?

1. Given the following equation. 2 Fe2O3 + 3 C → 4 Fe + 3 CO2 How many moles of carbon are required to completely react with 15.2 moles of iron(II)oxide?

2. Methane (CH4) burns completely with oxygen. How many moles of carbon dioxide will be produced when 21.3 moles of CH4 burns.

3. How many grams of sodium are in 34.2 moles of sodium?

I can't seem to figure out how to do these, if you could explain/show work that'd be great!

Answer 1

1) 22.8 moles
2) 15.2 moles
3) 786.6 grams

Explanation:

1) You just have to use mole ratio, which is obtained when you balance the equation.

`color(red)(2)Fe_2O_3 + color(red)(3)C -> 4Fe + 3CO_2`

Here you cross multiply to find the missing value,
`2 : 3`
`15.2` moles : `x`
`X= 22.8`

2) Same applies here, use mole ratio,

`color(red)(1)CH_4 + 2O_2 -> color(red)(1)CO_2 + 2H_2O`

Here you cross multiply to find the missing value,
`1 : 1`
`21.3 : X`
`x= 21.3`

3) You use the formula given,
Moles=Mass/Molar mass which is also;

`n=M/(Mr)`
So, you substitute in your values,
`34.2 =M/(23)`,

`M= 786.6`

Hope this helps!!

Answer 2

We gots `2Fe_2O_3(s) + 3C(s) rarr 4Fe(s) + 3CO_2(g)`

Explanation:

And so `"1.5 equiv of carbon"` are required to react with each `"equiv of ferric oxide"`...we gots `15.2*mol` ferric oxide `"(i.e. not iron(II) oxide)"`...and so we need `15.2*molxx3/2=22.8*mol*C`... an approx. mass of `275*g`.

For methane combustion, we should also write the stoichiometric equation...

`CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O`

And thus there is 1:1 stoichiometry between methane and carbon dioxide.... if we burn `21.3*mol` methane, we gets `21.3*mol` carbon dioxide...a mass of ...............

`21.3*molxx44.01*g*mol^-1~=940*g`

Sodium metal has a molar mass of `22.99*g*mol^-1`. How did I know this? If we gots a `34.2*mol` quantity, we gots a mass of...

`34.2*molxx22.99*g*mol^-1=??*g`