A hydrocarbon containing C and H has $92.3 \% \mathrm{C}$. When 39 g of hydrocarbon was completely burnt in…

A hydrocarbon containing C and H has $92.3 \% \mathrm{C}$. When 39 g of hydrocarbon was completely burnt in $\mathrm{O}_2, x$ moles of water and y moles of $\mathrm{CO}_2$ were formed. $x$ moles of water is sufficient to liberate 0.75 moles of $\mathrm{H}_2$ with Na metal. What is the weight (in g ) of oxygen consumed? $(\mathrm{C}=12 \mathrm{u} ; \mathrm{H}=1 \mathrm{u})$
  1. $120$
  2. $240$
  3. $360$
  4. $480$

Solution


empirical Ratio $1: 1$ CH possible hydrocarbon $\mathrm{C}_2 \mathrm{H}_2$ $\mathrm{C}_2 \mathrm{H}_2+\frac{5}{2} \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$ $2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{Na} \longrightarrow 2 \mathrm{NaOH}+\mathrm{H}_2$ 2 mole of $\mathrm{H}_2 \mathrm{O}$ gives one $\mathrm{mol} \mathrm{H}_2$ $0.75 \mathrm{H}_2$ Produced thus $1.5 \mathrm{~mol} \mathrm{H}_2 \mathrm{O}$ used mass of $1.5 \mathrm{~mol} \mathrm{H}_2 \mathrm{O}=1.5 \times 18=17$ On balancing and multiplying with 1.5 $\begin{gathered}1.5 \mathrm{C}_2 \mathrm{H}_2+3.75 \mathrm{O}_2 \longrightarrow 3 \mathrm{CO}_2+1.5 \mathrm{H}_2 \mathrm{O} \\ \text { mass of } \mathrm{O}_2 \text { required }=3.75 \times 32=120 \mathrm{~g}\end{gathered}$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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