HCl(aq) + NaOH(aq) ® NaCl(aq) + H2O(l) 50.0 mL of 1.00 M HCl are added to 50.0 m

Question

HCl(aq) + NaOH(aq) ® NaCl(aq) + H2O(l) 50.0 mL of 1.00 M HCl are added to 50.0 mL of 1.00 M NaOH in a calorimeter Temp increases from 21.0 to 27.5°C. What is r H in kJ/mol, assuming no heat loss and that the specific heat of the solution is the same as for water? C s,H20 = 4.184 J g- 1 K - 1, r H2O = 1.00 g/mL iPad 6:08 PM Example HCl(aq) NaOH (aq) Nacl (aq) H2O(D 50.0 mLof 1.oo M HCl are added to 50.0 mL of 1.oo M NaOH in a calorimeter Temp increases from 21.o to 27.5°C. What is Ar H in kj/mol, assuming no heat loss and that the specific heat o the solution is the same as for water? 4.184 g 1 K 1, pH20 E 1.oo g/mL s,H20 F 90%

Explanation / Answer

Total mass = Vtotal * Density

Vtotal = 50+50 = 100 mL

D = 1 g/mL so

mass = 100 g

C = 4.184 J/gC

dT = 27.5-21 = 6.5 °C

so

Q = m*C*(Tf-Ti) = 100*4.184*(6.5) = 2719.6 J

so

Qlost = Qrxn

then

mmol = MV = 50*1 = 50 mmol = 50*10^-3 mol

HRxn = -Q/n = -2719.6 / ( 50*10^-3 ) = -54392 J/mol

HRxn = -54.392 kJ/mol

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