Respuesta :
Answer: 404.04 kJ.
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
moles of [tex]H_2S[/tex]
[tex]\text{Number of moles}=\frac{26.7g}{34.1g/mol}=0.78moles[/tex]
[tex]2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)[/tex] [tex]\Delta H=-1036kJ[/tex]
According to stoichiometry :
2 moles of [tex]H_2S[/tex] on burning produces = 1036 kJ
Thus 0.78 moles of [tex]H_2S[/tex] on burning produces =[tex]\frac{1036kJ}{2}\times 0.78=404.04[/tex]
Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.
