limiting reactant question

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limiting reactant question

Postby SHU25 » Wed 04 May, 2011 15:20

I need help in this limiting reactant question
Please show all steps.
Thanks :)

If 42.7 g Cr2O3 and 9.8 g of Al react completely, how much Cr is formed?
What is the limiting reactant?
How much reactant is leftover?

`Cr_2O_3(s) + 2Al(l) → 2Cr(l) + Al_2O_3(l)`
SHU25
 
Posts: 4
Joined: Wed 04 May, 2011 12:35

Re: limiting reactant question

Postby Dama » Wed 04 May, 2011 15:38

`Cr_2O_3(s) + 2Al(l) -> 2Cr(l) + Al_2O_3(l)`
first find the moles
`Step 1: m=42.7g M=152 mass`
`n=m/M`
`n=2089mol`
`Step 2: n=m/M=9.8/27=0.368mol`
`Step 3: Cr_2O_3/0.281=2Cr/x=0.56mol`
`Step 4: 2Al/0.36=2Cr/x=0.36mol`
`Step 5: m=Mn=(52)(0.36)=18.72g`
`nexcess=ninitial=nreacted`
`0.281-0.18=0.101mol`
`m=Mn (52)(0.1)= 15.2g excess`
`nreacted 2Al/0.36=Cr_2O_3/x=0.18mol reacted`
Dama
 
Posts: 4
Joined: Wed 04 May, 2011 13:09


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