Question
How much water is produced from complete combustion of 30 gm propyne.
✓
Answer
$C _3 H _4+4 O _2 \longrightarrow 3 CO _2+2 H _2 O$
$(40 gm) \quad(2 \times 18 gm)$
According to balanced equation $H _2 O$ obtained from 1 mole $(40 gm )$ propyne $=2 \times 18= gm$
Water obtained from combustion of 30 gm propyne
$\begin{array}{l}
=\frac{2 \times 18 \times 30}{40} \\
=27 gm
\end{array}$
$(40 gm) \quad(2 \times 18 gm)$
According to balanced equation $H _2 O$ obtained from 1 mole $(40 gm )$ propyne $=2 \times 18= gm$
Water obtained from combustion of 30 gm propyne
$\begin{array}{l}
=\frac{2 \times 18 \times 30}{40} \\
=27 gm
\end{array}$
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