10cm3 of pure hydrocarbon was completely burnt in 95cm3 of oxygen (excess) at 425 k after cooling to room temperature,?
2018-02-28 6:20 pm
10cm3 of pure hydrocarbon was completely burnt in 95cm3 of oxygen (excess) at 425 k after cooling to room temperature, the gaseous mixture volume decreased from 125cm3 to 65cm3. A further reduction of 50cm3 was observed when the residual gas was passed through barium hydroxide. All gas volumes were measured at the same temperature and pressure. What is the formule of the hydrocarbon?
回答 (1)
2018-03-01 12:54 am
Let CₘHₙ be the formula of the hydrocarbon.
The balanced equation for the reaction at room temperature:
CₘHₙ(g) + [m + (n/4)]O₂(g) → mCO₂(g) + (n/2)H₂O(l)
According to the equation :
Volume of CₘHₙ reacted = 10 cm³
Volume of O₂ reacted = 10[m + (n/4)] cm³
Volume of CO₂ produced = 10m cm³
Volume of H₂O produced = 0 cm³
Volume of CO₂ produced (in cm³) :
10m = 50
m = 5
Volume of unreacted O₂ (in cm³) :
95 - 10[m + (n/4)] = 65 - 50
95 - 10m - 2.5n = 15
95 - 10×5 - 2.5n = 15
2.5n = 30
n = 12
The formula of the hydrocarbon = C₅H₁₂
The balanced equation for the reaction at room temperature:
CₘHₙ(g) + [m + (n/4)]O₂(g) → mCO₂(g) + (n/2)H₂O(l)
According to the equation :
Volume of CₘHₙ reacted = 10 cm³
Volume of O₂ reacted = 10[m + (n/4)] cm³
Volume of CO₂ produced = 10m cm³
Volume of H₂O produced = 0 cm³
Volume of CO₂ produced (in cm³) :
10m = 50
m = 5
Volume of unreacted O₂ (in cm³) :
95 - 10[m + (n/4)] = 65 - 50
95 - 10m - 2.5n = 15
95 - 10×5 - 2.5n = 15
2.5n = 30
n = 12
The formula of the hydrocarbon = C₅H₁₂
收錄日期: 2021-04-24 00:58:22
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