Respuesta :
Answer:
c
Explanation:
C4H10 + 13/2 O2 = 4CO2 + 5H2O
0.02 = 328*4.18*45
1= x
making x the subject of the formula you'll get 3084840.0J
Answer:
B) The enthalpy of combustion = 3,084,840 J
Explanation:
Given:
Moles of butane = 0.02
Mass of water, m = 328 g
Initial temperature T1 = 298 K
Final temperature T2 = 343 K
Specific heat of water, c = 4.18 J/g-K
To determine:
Enthalpy of combustion
Explanation:
Heat lost during combustion of butane = heat gained by water
Heat gained (q) by water is given as:
q = mc\Delta T = mc(T2-T1)
substituting for m, c, T2 and [tex]q = 328g*4.18J/g-K*(343-298)K = 61697.8 \ J[/tex]T1
[tex]Enthalpy \ of \ combustion = \frac{q}{moles\ of\ butane} \\\\= \frac{61697.8}{0.02} = 3,084,890\ J[/tex]
