Suppose 33 mL of 1.20 M HCl is added to 42 ml of 1.00 M NaOH in a coffee-cup cal
ID: 517371 • Letter: S
Question
Suppose 33 mL of 1.20 M HCl is added to 42 ml of 1.00 M NaOH in a coffee-cup calorimeter. The solution temperature, originally 25.0 degree C, rises to 31.8 degree C. Give the enthalpy change, delta H, for the reaction. HCl (aq0 + NaOH (aq) rightarrow NaCl (aq) + H_2 O (l) Assume the heat capacity and density of the final solution in the cup are those of water, 4.184 J/(g 0 degree C) and 1.00 g/mL. In an experiment, 7.36 g of zinc was heated with 6.45 g of sulfur. Assume that these substances react according to the equation 8Zn + S_8 rightarrow 8 ZnS. What amount of zinc sulfide was produced? How many grams of the excess reactant remain after the reaction is complete?Explanation / Answer
Volume of mixed solution =33+42= 75ml, since density is 1 g/ml, mass of solution = volume*density = 75*1 =75 gm
Enthalpy change = mass * specific heat* temperature difference= 75*4.184*(31.8-25)=2134 Joules
Moles of HCl in 33ml of 1.2M= Molarity* Volume(L)= 1.2*33/1000 =0.0396 moles
Moles of NaOH in 42ml of 1M= 1*42/1000 = 0.042 moles
Excess reactants is NaOH and limiting reactants is HCl. Hence molar enthalpy =2134/0.042 J/mole=50810 J/mole= 50.810 Kj/mole
2.
Atomic weight of Zinc = 65.38 and S= 32
Moles =mass/ atomic weight
Moles : Zinc= 7.36/65.38 =0.1125, S8= 6.45/(32*8)=0.0252
The reaction is 8Zn+S8----à 8ZnS
8 moles of Zinc requires 1 moles of S8. Molar ratio of Zn: S8= 8:1 ( stoichiometric ratio)
Actual ratio = 0.1125 :0.0252 = 4.5:1
So limiting reactant is Zn. Hence moles of ZnS formed = moles of Zn used = 0.1125
Molar mass of ZnS= 65.38+32= 97.38
Mass of ZnS= molar mass* moles of ZnS= 0.1125*97.38 =10.96 gm
Moles of S8 consumed =0.1125/8= 0.014
Moles of S8 remaining = 0.0252-0.014=0.0112
Mass of S8 remaining = molar mass* moles= 32*8*0.0112=2.87 gm
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