Sure, let's address each part of the problem step by step:
1. Determine the nature of the reaction:
- The reaction given is:
[tex]$2 Fe_2O_3(s) \rightarrow 4 FeO(s) + O_2(g) \quad \Delta H = 560 \text{ kJ}$[/tex]
- The value of [tex]\(\Delta H\)[/tex] is positive (560 kJ), indicating that the reaction absorbs energy.
- Therefore, the reaction is endothermic.
2. Determine whether heat will be released or absorbed when 56.3 g of [tex]\(Fe_2O_3\)[/tex] reacts:
- Since the reaction is endothermic (as established above), heat will be absorbed.
3. Calculate the amount of heat absorbed:
- First, we need to determine the molar mass of [tex]\(Fe_2O_3\)[/tex].
- The molar mass of [tex]\(Fe\)[/tex] is 55.845 g/mol, and the molar mass of [tex]\(O\)[/tex] is 15.999 g/mol.
- Therefore, the molar mass of [tex]\(Fe_2O_3\)[/tex] is:
[tex]$ \text{Molar mass of } Fe_2O_3 = 2 \times 55.845 + 3 \times 15.999 = 159.687 \text{ g/mol} $[/tex]
- Next, we calculate the number of moles of [tex]\(Fe_2O_3\)[/tex] in 56.3 g:
[tex]$ \text{Moles of } Fe_2O_3 = \frac{56.3 \text{ g}}{159.687 \text{ g/mol}} = 0.3525647047035763 \text{ moles} $[/tex]
- The given heat change ([tex]\(\Delta H\)[/tex]) is for 2 moles of [tex]\(Fe_2O_3\)[/tex], and [tex]\(\Delta H\)[/tex] for the reaction is 560 kJ. Therefore, the heat change per mole of [tex]\(Fe_2O_3\)[/tex] is:
[tex]$ \Delta H \text{ per mole} = \frac{560 \text{ kJ}}{2 \text{ moles}} = 280 \text{ kJ/mol} $[/tex]
- To find the heat change for the given moles of [tex]\(Fe_2O_3\)[/tex] (0.3525647047035763 moles), we multiply the moles by the heat change per mole:
[tex]$ \text{Heat change} = 0.3525647047035763 \text{ moles} \times 280 \text{ kJ/mol} = 98.718 \text{ kJ} $[/tex]
4. Final conclusion:
- The heat absorbed by the reaction when 56.3 g of [tex]\(Fe_2O_3\)[/tex] reacts is [tex]\( 98.718 \text{ kJ} \)[/tex].
Therefore, the detailed step-by-step solution is:
\begin{tabular}{|l|l|}
\hline This reaction is... & \textbf{endothermic.} \\
\hline Suppose 56.3 g of [tex]$Fe_2 O_3$[/tex] react. & \textbf{Yes, absorbed.} \\
\hline \begin{tabular}{l}
If you said heat will be released or absorbed in \\
the second part of this question, calculate how \\
much heat will be released or absorbed.
\end{tabular} & \\
\hline Round your answer to 3 significant digits. & [tex]\(\boxed{98.7 \text{ kJ}}\)[/tex] \\
\hline
\end{tabular}
