Question

You collected the following data in lab:

                          V_HCl = 50.15 mL         C_HCl = 0.960 M

                          V_NaOH = 48.90 mL     C_NaOH = 1.10 M

                          Tinitial ,HCl = Tinitial ,NaOH = Tinitial calorimeter = 25.7°C

                          T_f = 31.8°C

Assume that:

• S_NaOH = s_HCl = s_water = 4.18 J/g-°C
• C_cal = 20.5 J/^oC
• Solutions can be treated as dilute (i.e., physical properties such as density are approximately that of water)

1. Show one set of typed calculations for the following values: m_soln, T_soln, q_soln, q_neut, mole of the limiting reactant, and molar H_neut. Express molar H_neut in KJ/mol.

Can you show your work with an explanation so ill understand?

Answer 1

सध्या का आतिरिक्त कुछ fore write on this margin n = VⓇC limiting reactant NOH = Shu Iwater = 4.18 J 7- VHe= 50-15 ml Cha=0.960 moles He = 50.15 mx x 0.960 moles = 10.048 molest Jooo get X 1 "NaOH = 48.90 ml Canon = 1.40 notes n Nagh = 48.90 NX 1.10 moles 8 yon x 1.10 moles 10.05379 moles 1000 ad al Tinitial = 7; = 25.7°C T = 31.8°C NaOH + HCl Nad + H₂O + NaOH Секц) m=mass egy eech Atson - Tp-T; = 318 – 25.7°C 16.1°C n=no of moez 1 mole of Nach & I mole of Hill I mole of Nach so, d. 048 rules of Halk Nadu 10.046 mples of Nacht - M- Molar ass & 0.048 modes of H₂O. qeal - It or X Car = 61 cx 205 I = 125.05 |z. 0.048 moles Nach m=0.048 moles x 25 8.5 g mand = 2.808 g) 0.048 molls of H₂O MALO = 0.048 not x 189) to m=nT/ TM eelbed 158.59, m MERS 10.864

excess NaoH moles = 0.05379 -0.048 m = 5:079x103 moles. meras = 5.79 x 10 roles x 40 g 10.23169] since Speerific heat is same Potal sota mass = 2.8089 +0.864 9 +0.23169 Heat released neutralisation - absorbed by & Heat Heat calorimeter by soln sol = 3.9036g x 4.18 J x 6.1 c g-oc. Il soon = 99-5347 J Ineut - You + Pepu #25-08J +941534) I heut : +224.584 g +224.584 AH = - Inent no. of nodes = -224.584 J x kz 0.048 moles 1oooo Me