When \(30 \mathrm{~mL}\) of \(\mathrm{H}_2\) reacts with \(20 \mathrm{~mL}\) of \(\mathrm{O}_2\) to form…
- \(10 \mathrm{mLH}_2\)
- \(5 \mathrm{mLH}_2\)
- \(10 \mathrm{mLO}_2\)
- \(5 \mathrm{mLO}_2\)
Solution

Here, \(\mathrm{H}_2\) is the limiting reagent. So, \(30 \mathrm{~mL} \mathrm{H}_2\) consumes \(15 \mathrm{~mL}\) of \(\mathrm{O}_2\) as \(\mathrm{H}_2\) and \(\mathrm{O}_2\) react in \(2: 1\) volume ratio (at constant pressure and temperature).
Asked in: AP EAMCET 2020 (18 Sep Shift 2)
Practice more Some Basic Concepts of Chemistry questions on Aicharya