Given:I. 2Fe_{(s)} + frac{3}{2} O_{2(g)} to F | vedclass

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(A) To find the $\Delta H^{\Theta}$ for the reaction $3Mg_{(s)} + Fe_2O_{3(s)} 	o 3MgO_{(s)} + 2Fe_{(s)}$,we manipulate the given equations:Multiply

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$-227.2$

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(A) To find the $\Delta H^{\Theta}$ for the reaction $3Mg_{(s)} + Fe_2O_{3(s)} 	o 3MgO_{(s)} + 2Fe_{(s)}$,we manipulate the given equations:Multiply equation $(II)$ by $3$:$3Mg_{(s)} + \frac{3}{2} O_{2(g)} 	o 3MgO_{(s)}$; $\Delta H^{\Theta} = 3 	imes (-140.2) = -420.6 \ kJ$Reverse equation $(I)$:$Fe_2O_{3(s)} 	o 2Fe_{(s)} + \frac{3}{2} O_{2(g)}$; $\Delta H^{\Theta} = +193.4 \ kJ$Adding these two equations:$(3Mg_{(s)} + \frac{3}{2} O_{2(g)}) + (Fe_2O_{3(s)}) 	o (3MgO_{(s)}) + (2Fe_{(s)} + \frac{3}{2} O_{2(g)})$Canceling $\frac{3}{2} O_{2(g)}$ from both sides:$3Mg_{(s)} + Fe_2O_{3(s)} 	o 3MgO_{(s)} + 2Fe_{(s)}$$\Delta H^{\Theta} = -420.6 + 193.4 = -227.2 \ kJ$

Given:
$I$. $2Fe_{(s)} + \frac{3}{2} O_{2(g)} \to Fe_2O_{3(s)}$; $\Delta H^{\Theta} = -193.4 \ kJ$
$II$. $Mg_{(s)} + \frac{1}{2} O_{2(g)} \to MgO_{(s)}$; $\Delta H^{\Theta} = -140.2 \ kJ$
What is $\Delta H^{\Theta}$ of the reaction?
$3Mg_{(s)} + Fe_2O_{3(s)} \to 3MgO_{(s)} + 2Fe_{(s)}$
....... $kJ$

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Given:$I$. $2Fe_{(s)} + \frac{3}{2} O_{2(g)} \to Fe_2O_{3(s)}$; $\Delta H^{\Theta} = -193.4 \ kJ$$II$. $Mg_{(s)} + \frac{1}{2} O_{2(g)} \to MgO_{(s)}$; $\Delta H^{\Theta} = -140.2 \ kJ$What is $\Delta H^{\Theta}$ of the reaction?$3Mg_{(s)} + Fe_2O_{3(s)} \to 3MgO_{(s)} + 2Fe_{(s)}$....... $kJ$