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

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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)

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Gerda 1 week 2021-07-21T21:04:37+00:00 1 Answers 1 views 0

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    2021-07-21T21:06:17+00:00

    Answer:

    \Delta H=-113.3kJ

    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*\frac{1molC}{12.01gC} *\frac{1molCO_2}{1molC}=0.374molCO_2 \\\\9.21gO_2*\frac{1molO_2}{32.00gO_2} *\frac{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!

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