Question

The standard enthalpy of formation of CO2(g) is -393.5 kJ mol-1 . What is the enthalpy change if 4.49 g C(s) reacts with 9.21 O2(g) to form CO2(g)

Answer 1

`\Delta H=-113.3kJ`

Step-by-step explanation:

Hello there!

In this case, since the reaction for the formation of CO2 is:

`C+O_2\rightarrow CO_2`

Whereas the determination of the limiting reactant must be firstly performed as shown below:

`4.49gC*(1molC)/(12.01gC) *(1molCO_2)/(1molC)=0.374molCO_2 \\\\9.21gO_2*(1molO_2)/(32.00gO_2) *(1molCO_2)/(1molO_2)=0.288molCO_2`

Thus, we infer O2 is the limiting reactant because it yields less moles of CO2 in comparison to C; and therefore, the enthalpy change is:

`\Delta H=-393.5kJ/mol*0.288mol\\\\\Delta H=-113.3kJ`

Best regards!