The heat of formation of water is -260 kj/mol .How much water is decomposed by 130 kj of heat?
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Hey buddy,
● Answer- 9 g
● Explainaton-
# Given-
∆Hf = -260 kj/mol
∆H = 130 kj/mol
# Solution-
Heat of formation of water is -260 kj/mol thus it'll require 260 kj/mol to devompose.
No of molez = sample weight / molecular weight
n = x/18
Total heat required = No of moles × Heat of decomposition
130 = n × 260
n = 0.5
Putting values,
0.5 = x / 18
x = 9 g.
130 kj of heat will devompose 9 g of water.
Hope this is useful...
● Answer- 9 g
● Explainaton-
# Given-
∆Hf = -260 kj/mol
∆H = 130 kj/mol
# Solution-
Heat of formation of water is -260 kj/mol thus it'll require 260 kj/mol to devompose.
No of molez = sample weight / molecular weight
n = x/18
Total heat required = No of moles × Heat of decomposition
130 = n × 260
n = 0.5
Putting values,
0.5 = x / 18
x = 9 g.
130 kj of heat will devompose 9 g of water.
Hope this is useful...
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