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Gennadij [26K]
2 years ago
14

3 upper M n upper O subscript 2 (s) plus 4 upper A l (s) right arrow 2 upper A l subscript 2 upper O subscript 3 (g) plus 3 uppe

r M n (s). What is the enthalpy of the reaction?
Chemistry
1 answer:
Dvinal [7]2 years ago
5 0

<u>Answer:</u> The enthalpy of the reaction is -1791.31 kJ.

<u>Explanation:</u>

Enthalpy change is the difference between the enthalpies of products and the enthalpies of reactants each multiplied by its stoichiometric coefficients. It is represented by the symbol Delta H^o_{rxn}

\Delta H^o_{rxn}=\sum (n \times \Delta H^o_{products})-\sum (n \times \Delta H^o_{reactants})        .....(1)

For the given chemical reaction:

3MnO_2(s)+4Al(s)\rightarrow 2Al_2O_3(s)+3Mn(s)

The expression for the enthalpy change of the reaction will be:

\Delta H^o_{rxn}=[(2 \times \Delta H^o_f_{(Al_2O_3(s))}) + (3 \times \Delta H^o_f_{(Mn(s))})] - [(3 \times  \Delta H^o_f_{(MnO_2(s))}) + (4 \times \Delta H^o_f_{(Al(s))})]

Taking the standard heat of formation values:

\Delta H^o_f_{(Al_2O_3(s))}=-1675.7kJ/mol\\\Delta H^o_f_{(Al(s))}=0kJ/mol\\\Delta H^o_f_{(MnO_2(s))}=-520.03kJ/mol\\\Delta H^o_f_{(Mn(s))}=0kJ/mol

Plugging values in the above expression:

\Delta H^o_{rxn}=[(2 \times (-1675.7))+(3 \times 0)] - [(3 \times (-520.03))+(4 \times 0)]\\\\\Delta H^o_{rxn}=-1791.31 kJ

Hence, the enthalpy of the reaction is -1791.31 kJ.

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How many moles of chlorine gas at 120. °C<br> and 33.3 atm would occupy a vessel of 11.5<br> L?
Sever21 [200]

Answer:

11.9 moles Cl₂

Explanation:

To find the number of moles, you need to use the Ideal Gas Law. The equation looks like this:

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-----> V = volume (L)

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-----> R = constant (0.0821 L*atm/mol*K)

-----> T = temperature (K)

Before you can plug the given values into the equation, you first need to convert Celsius to Kelvin.

P = 33.3 atm                         R = 0.0821 L*atm/mol*K

V = 11.5 L                              T = 120. °C + 273.15 = 393.15 K

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(33.3 atm)(11.5 L) = n(0.0821 L*atm/mol*K)(393.15 K)

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