When 2.16 L of 1.800M Na_2S0_4 solution at 30.0degreeC is mixed with 1.42 L of 0

Question

When 2.16 L of 1.800M Na_2S0_4 solution at 30.0degreeC is mixed with 1.42 L of 0.360 M Ba(N0_3)_2 solution at 30.degreeC in a calorimeter, the white solid BaS0_4 forms, and the temperature mixture increases to 42.0degreeC. The overall balanced reaction is: Na_2S0_4(aq) + Ba(Na_3) rightarrow (aq) BaSO_4(s) + 2 NaNO_3(aq) a. Calculate the moles of BaSO_4(s) that precipitate, assuming the reaction goes to completion. b. Assuming that the calorimeter does not absorb any heat, the specific heat capacity of the solution is 6.37 J/*Cg, and the density of the final solution is 2.0 g/ml, calculate the enthalpy change per mole of BaS0_4 formed.

Explanation / Answer

The balance reaction is as follows:

Na2SO4 + Ba(NO3)2 -- > BaSO4 + 2 NaNO3

Now calculate the moles of Na2SO4 and Ba(NO3)2:

Moles of Na2SO4: 2.16 L* 1.800 Moles/ L= 3.888 Moles

Moles of Ba(NO3)2:1.42L*0.360 Moles/ L= 0.5112 Ba(NO3)2

Here Ba(NO3)2 is a limiting agent presence in small quantity and completely reacted in reaction .

0.5112 mol Ba(NO3)2* 1.0 mol BaSO4/1.0 mol Ba(NO3)2

= 0.5112 mol BaSO4

Hence there are 0.5112 mole BaSO4 formed.

q=mC(change in temp)
Total volume = 2.16+ 1.42 = 3.58 L or 3580mL

given that Density = 2.0 g/ mL

d= mass/ volume ;

mass = d* volume

mass = 2.0 g/ mL*3580mL

Mass= 7160 g

Heat capacity = 6.37 J / 0C –g

Temp. change = 42.0-30.0= 12.0 0C

Now calculate the total heat :

q=mC(change in temp)

q=7160 g *6.37 J / 0C –g* 12.0C

q= 547310.4 J

q= 547.3 KJ

Now we have 0.5112 mol BaSO4.
so divide heat by number of moles to calculate the enthalpy per mole of BaSO4 formed:

547.3 KJ / 0.5112 mol = 1070.64 KJ/mol

Hence the enthalpy per mole of BaSO4 formed is 1070.64 KJ/mol.

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