HC the whan 26 0 ) f K CO, was mixed with 30.0 ml of approximately 2M 2. When 2.
ID: 1029842 • Letter: H
Question
HC the whan 26 0 ) f K CO, was mixed with 30.0 ml of approximately 2M 2. When 2.76 g (0.0200 (a) Write a balanced equation for this reaction. (b) Calculate the enthalpy change (A H) of this reaction per mole of potassium (c) Explain WHY the hydrochloric acid need only be about 2 mol/L arbonate. Assume that the specific heat of the final mixture is 4.184 J/g°C, and that its density is 1.00 g/mL mixed with 3. Wh 30.0m 2 00g (0 0200 mole) of potassium hydrogen carbonate (KHCO) 30.0 ml of the same hydrochloric acid, the temperature falls by 3.7°C, (a) Write a balanced equation for this reaction. (b) Calculate the enthalpy change ( H) of this reaction per mole of K 4, wh en KHCO3 is heated, it decomposes into potassium carbonate, water, and carbon ide. By applying Hess's law and using the results from #2 and #3, calculate the enthalpy for the thermal decomposition of potassium hydrogen carbonate. Show all work and explain. Carefully note whether enthalpy changes are exothermic or endothermic.Explanation / Answer
2
a)
K2CO3 + HCl = H2O + CO2 + KCl
balance
K2CO3 + 2HCl = H2O + CO2 + 2KCl
b)
HRxn given:
Qsolution = m*C*(Tf-Ti)
Qsoln = 30*4.184*5.2 = 652.704 J
mol of K2CO3 = 0.02
mol of HCl = MV = (2)(30*10^-3) = 0.06
ratio is 0.02 --> 0.04, so we have excess HCl
HRxn = - Qsoln / mol of HCl reacted
= -652.704 /0.04
= -16317.6 J/mol
= -16.318 kJ/mol
c
because the limiting reagent is K2CO3
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