MHT CET · Chemistry · Some Basic Concepts of Chemistry
What amount of oxygen is used at S.T.P to obtain \(9 \mathrm{~g}\) water from sufficient amount of hydrogen gas?
- A \(5.6 \mathrm{dm}^3\)
- B \(22.4 \mathrm{dm}^3\)
- C \(16.8 \mathrm{dm}^3\)
- D \(11.2 \mathrm{dm}^3\)
Answer & Solution
Correct Answer
(A) \(5.6 \mathrm{dm}^3\)
Step-by-step Solution
Detailed explanation
\(
\mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(\ell)}
\)
\(11.2 \mathrm{dm}^3\) of \(\mathrm{O}_2\) gives \(18 \mathrm{~g}\) water at STP
\(\therefore 9 \mathrm{~g}\) water is obtained from \(\frac{112 \times 9}{18}=5.6 \mathrm{dm}^3\) of \(\mathrm{O}_2\)
\mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(\ell)}
\)
\(11.2 \mathrm{dm}^3\) of \(\mathrm{O}_2\) gives \(18 \mathrm{~g}\) water at STP
\(\therefore 9 \mathrm{~g}\) water is obtained from \(\frac{112 \times 9}{18}=5.6 \mathrm{dm}^3\) of \(\mathrm{O}_2\)
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