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Q.
Which of the following option represents correct limiting reagents in reactions (i), (ii) and (iii) respectively.
Some Basic Concepts of Chemistry
Solution:
$n_{ C }=\frac{26 \, g }{12 \,g / mol }=2.16$
$n_{ O _{2}}=\frac{20 \,g }{32 \, g / mol }=0.625$
$O _{2}$ will be a limiting reagent in reaction (i)
$n_{ N _{2}}=\frac{60 \,g }{28 \,g / mol }=2.14 $
$n_{ H _{2}}=40$
According to balanced equation,
$1 \,mol$ of $N _{2}$ requires $3$ mole of $H _{2}$
$2.14 \,mol$ of $N _{2}$ require $6.42 \,mol$ of $H _{2}$
$N _{2}$ will be a limiting reagent in reaction (ii)
$n_{ P _{4}}=\frac{100 \,g }{4 \times 31}=0.86 ; n_{ O _{2}}=6.25$
According to balanced equation
$1 \,mol$ of $P _{4}$ require $3 \, mol$ of $O _{2}$
$0.86 \,mol$ of $P _{4}$ require $2.58 \, mol$ of $O _{2}$
So, $P _{4}$ is a limiting reagent in reaction (iii)
