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Physical Chemistry

Cool goIITian

 Joined: 23 Apr 2012 Post: 36
8 Jun 2012 19:17:53 IST
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Stiochiometry and limiting reagents
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In the reaction

Al4C3  +  12H2→ 4Al(OH)3  +  3CH4

What would be the  minimum mass of Al4C3 required to obtain atleast 50g of Al(OH)3 and 50g of methane?

please explain how u got the answer ? it will be very nice of u !!!!!!!!

#### Comments (4)

Scorching goIITian

Joined: 5 May 2012
Posts: 217
8 Jun 2012 21:00:43 IST
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Al4C3  +  12H2→ 4Al(OH)3  +  3CH4

Clearly, mass of Al4C3 will be determined according to methane a its required mole are more (50/16 to be precise). Now, 48 gm of methane is formed for every 144gm  of Al4C3. => (50*144)/48 gm is required for 50 gm of methane. Hence, minimum mass reqd = 150gm. Like if found useful.

Cool goIITian

Joined: 23 Apr 2012
Posts: 36
8 Jun 2012 22:10:49 IST
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thnx dude ur info was very damn useful thnx for ur help

Forum Expert
Joined: 29 Dec 2006
Posts: 5524
9 Jun 2012 15:33:27 IST
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mol weight of Al4C3 = 4*27+3*12 = 144. So 144 g Al4C3 gives 16 g methane. For 50 g methane we need (50/16)*144 g Al4C3

Forum Expert
Joined: 29 Dec 2006
Posts: 5524
9 Jun 2012 15:34:37 IST
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Sorry for the mistake. 144 g Al4C3 gives 3 moles = 3*16 = 48 g methane. So for 50 g we need (50/48)*144 g Al4C3 = 150 g

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