Use the data in the following image to answer this question.a.The reaction involved the neutralization of xxxmol of HCl, therefore, calculate the molar heat of neutralization. DETERMINATION OF AHneu FOR HCIIt is essential to keep track of which quantitites are actually measuredin this exercise and which ones are given.Suppose we mix samples of measured volumes and measuredtemperatures and given concentrations of HCI and NaOHVNaOH =VHI = 50.0 mL55.0 mLTNGOH =24.2°CTна 24.0°CMNaOH =1.30MMHa = 1.20 MWe measure temperature change at t = 0 graphically. The result isthat AT = 4.75C° .We need to know how much of the essentialreactant (HCI) has been consumed. That requires determining thelimiting reagent.Using the measured volumes and given concentrations of HCl andNAOH, we can calculate the number of moles of each reactant.NHCI= 0.0600 mol%3DМнаNNOOH =VNOOH *= 0.0715 molMNeOHThis makes HCI the limiting reagent. Only 0.0600 mol HCl can beneutralized.Using the given heat capacity, given densities and measured volumes ofthe solutions and the measured temperature change, we can analyzethe heat exchange, q, and correct for the heat loss to the calorimeterusing the CalorimetricCcal, alreadetermined.Heat Capacity = 3.97 J/gDensity = 1.02 g/mLCcal = 42.0 J/C°Tot Vol of Sol = 50.0 + 55.0 = 105.0 mL

Given: HCl is the limiting reagent. Moles of HCl =0.0600 mol ΔT =4.75oC Heat capacity = 3.97 J/g…

Answered: Use the data in the following image to…

Use the data in the following image to answer this question. a.The reaction involved the neutralization of xxxmol of HCl, therefore, calculate the molar heat of neutralization.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

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Question

Use the data in the following image to answer this question.

a.The reaction involved the neutralization of xxxmol of HCl, therefore, calculate the molar heat of neutralization.

DETERMINATION OF AHneu FOR HCI
It is essential to keep track of which quantitites are actually measured
in this exercise and which ones are given.
Suppose we mix samples of measured volumes and measured
temperatures and given concentrations of HCI and NaOH
VNaOH =
VHI = 50.0 mL
55.0 mL
TNGOH =
24.2°C
Tна 24.0°C
MNaOH =
1.30M
MHa = 1.20 M
We measure temperature change at t = 0 graphically. The result is
that AT = 4.75C° .We need to know how much of the essential
reactant (HCI) has been consumed. That requires determining the
limiting reagent.
Using the measured volumes and given concentrations of HCl and
NAOH, we can calculate the number of moles of each reactant.
NHCI
= 0.0600 mol
%3D
Мна
NNOOH =
VNOOH *
= 0.0715 mol
MNeOH
This makes HCI the limiting reagent. Only 0.0600 mol HCl can be
neutralized.
Using the given heat capacity, given densities and measured volumes of
the solutions and the measured temperature change, we can analyze
the heat exchange, q, and correct for the heat loss to the calorimeter
using the Calorimetric
Ccal, alrea
determined.
Heat Capacity = 3.97 J/g
Density = 1.02 g/mL
Ccal = 42.0 J/C°
Tot Vol of Sol = 50.0 + 55.0 = 105.0 mL

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