A 72.0 mL sample of 1.00 M NaOH is mixed with 59.0 mL of...

A 72.0 mL sample of 1.00 M NaOH is mixed with 59.0 mL of 1.00 M H2SO4 in a large Styrofoam coffee cup; the cup is fitted with a lid through which passes a calibrated thermometer. The temperature of each solution before mixing is 20.4°C. After adding the NaOH solution to the coffee cup, the mixed solutions are stirred until reaction is complete. Assume that the density of the mixed solutions is 1.00 g/mL, that the specific heat of the mixed solutions is 4.184 J/(g °C), and that no heat is lost to the surroundings. The ?Hrxn for the neutralization of NaOH with H2SO4 is – 114 kJ/mol H2SO4. What is the maximum measured temperature (in °C) in the Styrofoam cup? [Report your answer to 1 decimal place, do not include the units].

Transcript

00:01 Hey welcome to the video, we have the reaction of naoh and h2so4 and this gives na2so4 plus h2o and heat where the heat given is calculated with the formula q is equal to m into c into delta t and for this we can calculate the m to be equal to the volume into the density given where the is equal to the volume of naoh which is 72 .0 ml plus the volume of h2so4 added which is 59 .0 ml and this would be equal to 131 .0 ml then from this we would have the mass to be equal to 131 .0 ml into 1 .00 g per ml and this would be equal to 131 .0 g and we know that delta h is equal to q divided by the number of moles here the number of moles depends upon the limiting reagent and over here the number of moles of the reactants naoh is equal to 0 .072 liters into 1 .00 mole per liter which would be equal to 0 .072 mole and the number of moles of h2so4 needed is equal to half of this which is 0 .072 mole divided by 2 which would be equal to 0 .036 mole while the number of moles of h2so4 present is equal to its liter which is 0 .059 liters into 1 .00 mole per liter which would be equal to 0 .059 mole here as the number of moles of h2so4 present is greater than the number of moles of h2so4 needed the number of moles of naoh is the limiting reagent and therefore the number of moles of the reactants is equal to 0 .072 mole then from this we would have q to be equal to the delta h value which is 114 into 10 to the power 3 joule per mole divided by mole into the number of moles which is 0 .072 mole and from this we would have the value of q to be equal to 8 .208 joules and therefore from this on substituting the values we would have it to be equal to 8 .208 into 10 to the power 3 joule to be equal to the mass which is 131 .0 gram into the specific heat capacity which is 4 .184 joule per gram degree celsius into tf minus 20 .4 degree celsius so therefore from this we would have tf minus 20 .4 to be equal to 1 .497 degree celsius and therefore from this we would have tf to be equal to 1 .497 degree celsius plus 20 .4 degree celsius and the final temperature that is tf is equal to 21 .897 degree celsius...

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