A 101.2mL sample of 1.00MNaOH is mixed with 50.6 mL of 1.00 MH SO4in a large Sty

Question

A 101.2mL sample of 1.00MNaOH is mixed with 50.6 mL of 1.00 MH SO4in a large Styrofoam coffee cup: the cup is fitted with a lid through whiclh passes a calibrated thermometer. The temperature of each solution before mixing is 21.45 °C. After adding the NaOH solution to the coffee cup and stirring the mixed solutions with the thermometer, the maximum temperature measured is 31.30 Assume that the density of the mixed solutions is 1.00 g/mL, that the specific heat of the mixed solutions is 4.18 JsC), and that no heat is lost to the surroundings 2nd attempt Part 1 (2 points) hi See Periodic Table Write a balanced chemical equation for the reaction that takes place in the Styrofoam cup. Remember to include phases in the balanced chemical equation. 2NOH (aq) + H2S04- Na2S04 (aq) + 2H20(1)

Explanation / Answer

Part 1.
2NaOH (aq) + H2SO4(aq) ----> Na2SO4 (aq) + 2H2O(l)

Part 2.
No. moles of NaOH = Molarity x volume in L
                  = 1.0 mol/L x 0.1012 L
                  = 0.1012 mol
No. moles of H2SO4= Molarity x volume in L
                  = 1.0 mol/L x 0.0506 L
                  = 0.0506 mol
              
From the stoichiometry, 2 mol of NaOH reacts with 1 mol of H2SO4. Therefore, 0.1012 mol of NaOH requires 0.0506 mol
of H2SO4. So, there will be competing utilization of NaOH and H2SO4 and produces Na2SO4 and H2O.

SO, There is no NaOH and H2SO4 left in the cup, hence, the answer is NO.

Part 3.
The heat of neutralization is:
Q = mcT,

Given that volume of NaOH + H2SO4 = 101.2 mL + 50.6 mL = 151.8 mL

mass = mass of the solution = volume x density = 151.8 mL x 1.0 g/mL = 151.8 g
T1 = 21.45 oC and T2 = 31.3 oC

T = T2-T1 = 31.3 - 21.45 = 9.85 oC

The heat of neutralization is:
Q = 151.8 g x 4.18 J/g.oC x 9.85 oC
= 6250 J

molar enthalpy of reaction H = Q/n

n = no. moles of H2SO4. = 1

Therefore, H = 6250 J/1mol = 6250 J/mol = 6.250 KJ/mol(H2SO4)

Hence, the answer is = 6.250 KJ/mol(H2SO4)

                 

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