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Physical Chemistry

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8 Jun 2012 19:17:53 IST

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Stiochiometry and limiting reagents

Engineering Entrance , Medical Entrance , AIPMT , JEE Main , AIIMS , JEE Advanced , Chemistry , Physical Chemistry

In the reaction

Al₄C₃ + 12H₂O → 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)

Uppinder Chugh

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Joined: 5 May 2012

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8 Jun 2012 21:00:43 IST

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Al₄C₃ + 12H₂O → 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.

Ameya Uddhav Harmalkar

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8 Jun 2012 22:10:49 IST

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thnx dude ur info was very damn useful thnx for ur help

Krishna Gopal Singh

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Joined: 29 Dec 2006

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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

Krishna Gopal Singh

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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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