A. Carbon (II) oxide
B. Carbon (IV) oxide
C. Oxygen
D. Nitrogen
Correct Answer:
Option C – Oxygen
Explanation
2CO + O2 → 2CO2
50cm³ → 25cm³
20% X 150 = 30cm³
30cm³ is the volume of oxygen available for the reactions. But only 25 cm³ is required; therefore O2 is in excess
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![A given volume of oxygen diffused through a porous partition in 8 seconds. How long would it take the same volume of carbon (IV) oxide to diffuse under the same condition? [C = 12.0, O = 16.0]](https://erudites.ng/wp-content/plugins/contextual-related-posts/default.png "A given volume of oxygen diffused through a porous partition in 8 seconds. How long would it take the same volume of carbon (IV) oxide to diffuse under the same condition? [C = 12.0, O = 16.0]")
