You have reacted 100.00mL of 1.353M aqueous sulfuric acid with 12.618g of sodium hydroxide solid. If all of the heat generated by this reaction is transferred to a 1.317kg block of copper metal initially at 16.82°C, what is the final temperature of the block of copper metal? (Specific heat of copper = 0.375J/g*°C)

70.137 °C

Explanation:

The reaction generated from the question can be expressed as:

The enthalpy reaction:

Repacing the values of each compound at standard enthalpy conditions;

no of moles of NaOH = 12.618g/39.99 g/mol

= 0.3155 mol

no of moles of H₂SO₄ = molarity of H₂SO₄ × Volume

= 1.3553 mol/L × 100 × 10⁻³ L

= 0.13553 mol

From the reaction,

1 mol of NaOH = 2 × mol of H₂SO₄

Since mol of NaOH is greater than that of  H₂SO₄, then NaOH is the excess reagent and H₂SO₄ is the limiting reactant

∴

1 mol of H₂SO₄ yields = - 194.29 kJ

0.13553 mol of H₂SO₄ will yield;

= -26.332124 kJ

= -26332.12 J

Finally,

Heat(q) =

26332.12 J = 1.317 × 10³ g × 0.375 J/g°C ×ΔT

26332.12 J = 493.875 J/° C  × ΔT

26332.12 /  493.875 = ΔT

ΔT = 53.317 °C

- 16.82 °C = 53.317 °C

= (53.317 + 16.82) °C

= 70.137 °C

i'm assuming that it's c.) red bank..

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2