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Chemistry: Post your doubts here!

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I am not too sure about this but I think D is the right answer because as you down the alkane homologous series, the boiling point and melting point increases. So, maybe more energy would be needed for larger alkanes to undergo combustion and therefore more residual gas is produced by smaller alkanes over a certain period of time.
 
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Please help with these...i will really appreciate it!
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Ans B

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1 and 2 correct
 
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U dont even have 2 do any calculations of da stoichiometry. Valency of C is 4, valency of Al is 3, so u cross multiply n forms Al₄C₃. All other options wud fail 2 satisfy da octet rule

S11Q12

In this case, the cross multiplication does give us the answer, but have to be careful at A levels, not all species need to satisfy the octet rule.
E.g AlCl3 and SO4 2-


This was what i suggest some posts ago

Understand that whatever moles of C we have at the beginning will eventually be the same number of moles of C in CO2 at the end.

Let x and y be the ratio of Al and C respectively.
AlxCy --->???---> CO2

Working backwards from CO2,
moles of CO2 = 72/24 000 = 0.003 moles = moles of C in AlxCy

Going through the options
A Al 2C3, moles of Al2C3 = 0.144/90 = 0.0016, moles of C = 0.0016 x 3 = 0.0048 (incorrect)
B Al 3C4 , moles of Al3C4 = 0.144/129 = 0.0016, moles of C = 0.0016 x 4 = 0.0045 (incorrect)
C Al 4C3 , moles of Al4C3 = 0.144/144 = 0.001, moles of C = 0.001 x 3 = 0.003 (BINGO!)
D Al 5C3
 
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