Question

A. How many kilocalories are released when 215 g Cl2 reacts with silicon?

B. Assuming that Coca-Cola has the same specific heat as water [4.18 J/(g⋅∘C)], calculate the amount of heat in kilojoules transferred when one can (about 350 g) is cooled from 30 ∘C to 8 ∘C.

C. When 25.0 mL of 1.0 M H2SO4 is added to 50.0 mL of 1.0 M NaOH at 25.0 ∘C in a calorimeter, the temperature of the aqueous solution increases to 33.9 ∘C.

Assuming that the specific heat of the solution is 4.18 J/(g⋅∘C), that its density is 1.00 g/mL, and that the calorimeter itself absorbs a negligible amount of heat, calculate ΔH in kilojoules for the reaction.

H2SO4(aq)+2NaOH→2H2O(l)+Na2SO4(aq)

Answer 1

A) The reaction is--- Si (s) + 2Cl₂ (g) →----> SiCl₄ (s) , H = -157 kcal

Molar mass of Cl₂ = 71 g/mol

215 g Cl₂ x 1mol Cl₂/71 g Cl₂ = 3.03 mol Cl₂

3.03 mol Cl₂ x (-157 kcal/2mol Cl₂ ) = -237.85 kcal

237.85 kilocalories are released.

B) q = m x c x T

=> q = 350 g x 4.18 J/g⁰C x (8-30)⁰C

=> q = (350 x 4.18 x -22) J

=> q =- 32186 J =-32.186 KJ

C) The given reaction is, H₂SO₄(aq) + 2NaOH-----> 2H₂O(l) + Na₂SO₄(aq)

mass = density x volume = 1.00g /mL x (25 +50) mL = 1.00 g/mL x 75 mL = 75 g

T = (33.9 -25.0)⁰C = 8.9⁰C

Now,

q = mcT

=> q = 75 g x 4.18 J/g⁰C x 8.9⁰C

=> q =2790.15 J

Again,

(0.0250 L) x (1.0 mol/L H₂SO₄) = 0.025 mol H₂SO₄

Now,

H =-q/0.025 mol = -(2790.15 J) / (0.025 mol H₂SO₄) = -111606 J/mol = -111.6 kJ/mol H₂SO₄

Therefore,
H₂SO₄(aq) + 2 NaOH(aq) -----> 2 H₂O(l) + Na₂SO₄, H = -111.6 kJ