The warmth of neutralization of $ceHCl(aq)$ by $ceNaOH$ is $pu-55.9 kJ/mol$ $ceH2O$ produced. If $pu50ml$ the $pu1.6M$ $ceNaOH$ at $pu25.15^circ C$ is included to $pu25ml$ of $pu1.79M$ $ceHCl$ at $pu26.34^circ C$ in a plastic foam cup calorimeter, what will be systems temperature be immediately after the neutralization reaction has occurred?

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 Buck Thorn♦
request Nov 15 "14 at 2:57 JackJack
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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 Buck Thorn♦
answer Nov 17 "14 in ~ 20:48 DavePhDDavePhD
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