I"m always overwhelmed through the inquiry if it provided me too much information. Ns don"t recognize what equation and also concept to use first. (But I understand all related-equations in this unit.)
I require a thorough explanation to understand! I"m not certain which one is the early stage temperature? $pu25.15^circ C$ the $ceNaOH$ or $pu26.34^circ C~ceHCl$?
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edited might 19 in ~ 6:24
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First, you have to recognize what the limiting reagent is. Because NaOH and also HCl reaction in a 1:1 ratio, and there is less HCl ($pu25mL$ the $pu1.79M$ is under moles 보다 $pu50 mL$ the $pu1.6M$), HCl is the limiting reagent.
You are watching: How to calculate heat of neutralization
Second, offered the amount of limiting reagent, just how much warm will be exit (or absorbed). $pu0.025L imes pu1.79M imes pu55.9 kJ/mol$
Third, you have to approximate the the solution has the warmth capacity the water, i beg your pardon is $pu4.18 kJ/K L$.
If girlfriend mix 2 volumes that the very same substance at different temperatures, the temperature of merged volumes will certainly be approximately the volume-weighted average. Therefore here: $$(50 imes 25.15 + 25 imes 26.35)/75 = 25.55$$
So lastly calculate the quantity the temperature the $pu75 ml$ of "water" will increase when $pu0.025L imes pu1.79M imes pu55.9 kJ/mol$ of heat is added, and add this come $pu25.55 ^circ C$.
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edited may 19 at 6:27
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answer Nov 17 "14 in ~ 20:48
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