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An excess of zinc metal is added to $50.0 \mathrm…
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Problem 114 Medium Difficulty

Calculate the $\Delta H^{\circ}$ for the reaction $$\mathrm{Fe}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{3}(s)$$

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Video Transcript

So for this question, were supposed to find Delta H for the following chemical reaction. And so it's always the strategy in finding the Delta. H is going to be finding the heat of formation products and then subtracting out the heat information of the reactive so we can express it as it'll be the heat information product minus the heat information of reactions. In this case, it's pretty straightforward because we only have three compounds and and ions to deal with. So first we have one mole of F E. O. H. Three, so we would write one. And then we know that the heat of formation of F. E. O. H. Three is negative. 124 0.25 Q jewels from all. Then you need to subtract out the heat information for the F E three plus ion and the heat information. For that, it's negative. 47.7 killer jewels promote. Then finally, we need to once again subtract out three times because there's three moles of the hydroxide ion of the heat information that which is negative. 229 white, 94 killer jewels. More so when you plug all of that. And your calculator, you'll get Delta H for the entire reaction is negative. 86.7 kill jewels for more, which is the final answer.

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