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!

2 Answers
Mar 13, 2018

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!!

Mar 13, 2018

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