35 cm3 of hydrogen was sparked with 12 cm3 of oxygen at 110°C and 760mm Hg to produce steam. What percentage of the total volume of gas left after the reaction is hydrogen?
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Correct Answer: Option B
Explanation:
Reaction for the equation:
\(2H_{2} + O_{2} \to 2H_{2} O\)
\(2 vol : 1 vol : 2 vol\)
Hence, 2 volumes of Hydrogen is needed for the reaction with Oxygen to produce steam (water).
\(\therefore 12cm^{3} O_{2} \to 2 \times 12cm^{3} = 24cm^{3} H_{2}\)
\(\text{Excess of Hydrogen} : 35 cm^{3} - 24 cm^{3} = 11 cm^{3}\)
\(\text{% Hydrogen left after the reaction} = \frac{11}{35} \times 100% = \frac{220}{7} = 31.48% \)
\(\approxeq 31% \)
Reaction for the equation:
\(2H_{2} + O_{2} \to 2H_{2} O\)
\(2 vol : 1 vol : 2 vol\)
Hence, 2 volumes of Hydrogen is needed for the reaction with Oxygen to produce steam (water).
\(\therefore 12cm^{3} O_{2} \to 2 \times 12cm^{3} = 24cm^{3} H_{2}\)
\(\text{Excess of Hydrogen} : 35 cm^{3} - 24 cm^{3} = 11 cm^{3}\)
\(\text{% Hydrogen left after the reaction} = \frac{11}{35} \times 100% = \frac{220}{7} = 31.48% \)
\(\approxeq 31% \)