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Amount of oxygen required for complete combustion of 27g Al is

$(a)\;24g\qquad(b)\;12g\qquad(c)\;20g\qquad(d)\;6g$

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$4Al+3O_2\rightarrow 2Al_2O_3$
$4\times 27g$ Al reacts with $3\times 32g O_2$
$\therefore 27g$ Al reacts with $\large\frac{3\times 32\times 27}{4\times 27}$
$\Rightarrow 24g\;O_2$
Hence (a) is the correct answer.
answered Jan 8, 2014 by sreemathi.v
 

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