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### Topic: Limiting reagents in a reaction  (Read 2503 times)

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#### myicc

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##### Limiting reagents in a reaction
« on: June 04, 2024, 07:41:17 AM »
In the reaction represented by 3 Fe + 4 H2O → Fe3O4 + 4 H2, how many moles of H2 and what mass of Fe3O4 will be formed by the reaction of 5.10 moles of Fe?

a) 4 moles of H2 and 232g of Fe3O4 formed
b) 6.8 moles of H2 and 394g of Fe3O4 formed
c) 6.8 moles of H2 and 1.7 g of Fe3O4 formed
d) 20.4 moles of H2 and 1.18kg of Fe3O4 formed

(I got 6.8 moles of H2 by: 3moles Fe divided 5.10moles Fe ×4moles H2 = 6.8 moles of H2 and 1.18kg of Fe3O4 by 5.10x231.55 (molar mass of Fe3O4) but obviously this is not an option so I'm somewhat confused)

#### mjc123

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##### Re: Limiting reagents in a reaction
« Reply #1 on: June 04, 2024, 09:00:38 AM »
Why 5.10 x 231.55? If you have 5.10 moles Fe, how many moles of Fe3O4 do you get?

#### GibbsFreeEnergy

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##### Re: Limiting reagents in a reaction
« Reply #2 on: July 30, 2024, 06:42:24 PM »
The correct answer is b) (n(H2)=6.8 mol, m(Fe3O4)=394 g)
Calculating n(H2):
n(H2)/n(Fe)=4/3
n(H2)=n(Fe)*4/3=5,10*4/3=6,8 mol
Calculating m(Fe3O4):
n(Fe3O4)/n(Fe)=1/3
n(Fe3O4)=n(Fe)*1/3=5,10*1/3=1,7 mol
m(Fe3O4)=n(Fe3O4)*M(Fe3O4)=1,7 mol*231,533 g/mol=393,6061 g

#### Corribus

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##### Re: Limiting reagents in a reaction
« Reply #3 on: July 30, 2024, 07:04:35 PM »
Although this post is pretty old, please don't just give answers to people. At Chemical Forums we like to encourage people to think through problems themselves. This is the only way they learn.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman