Asked by anon
Calculate the enthalpy of formation if 78.5 g of carbon dioxide in the following reaction:
C(s) + H2O(g) --> CO2(g)
Use the following equations:
a) H2O(l) --> H2(g) + (1/2)O2(g): Δ°f = +285.8 kJ/mol
b) C2H6(g) --> 2C(s) + 3H2(g): Δ°f = +84.7 kJ/mol
c) 2CO2(g) + 3O2(g) --> C2H6(g) + (7/2)O2(g): Δ°f = +1560.7kJ/mol
C(s) + H2O(g) --> CO2(g)
Use the following equations:
a) H2O(l) --> H2(g) + (1/2)O2(g): Δ°f = +285.8 kJ/mol
b) C2H6(g) --> 2C(s) + 3H2(g): Δ°f = +84.7 kJ/mol
c) 2CO2(g) + 3O2(g) --> C2H6(g) + (7/2)O2(g): Δ°f = +1560.7kJ/mol
Answers
Answered by
DrBob222
I don't believe you can do that.
C(s) + H2O(g) ==> CO2(g)
a. The equation is not balanced.
b. In those equations to be used you don't have H2O(g) anywhere but only H2O(l)
c. The equation you have posted isn't balanced. You're OK with C but O doesn't balance and there's no H on the left. Try again.
C(s) + H2O(g) ==> CO2(g)
a. The equation is not balanced.
b. In those equations to be used you don't have H2O(g) anywhere but only H2O(l)
c. The equation you have posted isn't balanced. You're OK with C but O doesn't balance and there's no H on the left. Try again.
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