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If 6.400 g of C6H6 is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final temperature of the water?

Respuesta :

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2C6H6(l)+15O2(g) ----> 12CO2(g) +6H2O(l) + 6542 kJ 

C6H12 = 6x12 + 6x1 = 78. 
The equation indicates that 2x78 = 156g benzene will produce 6542kJ. 
Using proportions you can then calculate that 
x/6542kJ = 6.4g / 156g 
x = 268.39 kJ = 268390 J 

heat = mass x ΔT x 4.18J/g° 
ΔT = 268390 J / (5691g x 4.18J/g°) = 11.28 °C

final temp = 21 + 11.28° = 32.28°C

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