WAEC - Chemistry (1999 - No. 21)
30cm3 pf hydrogen at s.t.p combine with 20cm3 of oxygen to form steam according to the following equation 2H2(g) + O2(g) → 2H2O(g).
Calculate the total volume of the gaseous mixture at the end of the reaction
50cm3
35cm3
30cm3
25cm3
Explanation
2H\(_2\)(g) + O\(_2\)(g) → 2H\(_2\)O(g).
2 volumes of H\(_2\)(g) and 1 volume of O\(_2\)(g) produced 2 volumes of H\(_2\)O\(_2\)
ratio of reacting volumes
30cm\(^3\) + 15cm\(^3\) → 30cm\(^3\)
30cm\(^3\) of Hydrogen requires 15cm\(^3\) of oxygen to produce 30cm\(^3\) of water.
Excess oxygen: 20cm\(^3\) – 15cm\(^3\) = 5cm\(^3\)
Total volume of gaseous mixture: 5cm\(^3\) of Oxygen + 30cm\(^3\) of Water = 35cm\(^3\)
2 volumes of H\(_2\)(g) and 1 volume of O\(_2\)(g) produced 2 volumes of H\(_2\)O\(_2\)
ratio of reacting volumes
30cm\(^3\) + 15cm\(^3\) → 30cm\(^3\)
30cm\(^3\) of Hydrogen requires 15cm\(^3\) of oxygen to produce 30cm\(^3\) of water.
Excess oxygen: 20cm\(^3\) – 15cm\(^3\) = 5cm\(^3\)
Total volume of gaseous mixture: 5cm\(^3\) of Oxygen + 30cm\(^3\) of Water = 35cm\(^3\)
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