Answer:
The final temperature of the mixture is 28.11 °C
Explanation:
Step 1: Data given
Volume of 1.00 M Ba(NO3)2 = 1.00 L
Temperature = 25.0 °C
Volume of 1.00 M Na2SO4 = 1.00 L
enthalpy change is – 26 kJ per mol BaSO4
The specific heat of water is 4.18 J/g ·˚C
the density of water is 1.00 g/mL
Step 2: The balanced equation
Ba(NO3)2(aq) + Na2SO4(aq) → 2NaNO3(aq) + BaSO4(s)
Step 3: Calculate the total volume
Total volume = 1.00 L + 1.00 L = 2.00 L = 2000 mL
Step 4: Calculate mass
Mass = volume * density
Mass = 2000 mL * 1g/mL
Mass = 2000 grams
Step 5: Calculate moles BaSO4 formed
For 1 mol Ba(NO3)2 we need 1 mol Na2SO4 to produce 1 mol BaSO4
There is no limiting reactant, both Ba(NO3)2 and Na2SO4 will be completely be consumed (1 mol). We'll have 1.0 mol of BaSO4 produced.
Step 6: Calculate Q
Q = - ΔH
ΔH is negative so the reaction is exothermic, what means the temperature increases
Q is always positive, so Q = 26kJ = 26000 J
Step 6: Calculate the heat transfer
Q= m*c*ΔT
⇒with Q = the heat transfer = TO BE DETERMINED
⇒with m =the mass of the solution = 2000 grams
⇒with c= the specific heat of the solution = 4.18 J/g°C
⇒with ΔT = the change of temperature = T2 - T1 = T2 - 25.0
26000 = 2000 * 4.18 * (T2 - 25.0 °C)
3.11 = T2 - 25.0 °C
T2 = 25.0 + 3.11 °C
T2 = 28.11 °C
The final temperature of the mixture is 28.11 °C
= 6.32 mol NO