Answer:
26. 42.3 kJ; 27. 143 kJ
Explanation:
26. Formation of water gas
We know that we will need the balanced equation with masses, molar masses, and enthalpies, so let’s gather all the information in one place.
A_r: 12.01
C + H₂O ⟶ CO + H₂; ΔH_r = +113 kJ
Mass/g: 4.50
(a) Calculate the moles of C
Moles of C = 4.50 × 1/12.01 = 0.375 mol C
(b) Calculate the energy absorbed
The conversion factor is 113 kJ/1 mol Al
Heat = 0.375 × 113/1 = 42.3 kJ
27. Oxidation of glucose
A_r: 180.16
C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O; ΔH_r = -2803 kJ
Mass/g: 9.22
(a) Calculate the moles of C₆H₁₂O₆
Moles of C₆H₁₂O₆ = 9.22 × 1/180.16 = 0.051 07 mol C₆H₁₂O₆
(b) Calculate the energy absorbed
The conversion factor is (-2803 kJ/1 mol C₆H₁₂O₆)
Heat = 0.051 07 × (-2803)/1 = -143 kJ
The oxidation of glucose releases 143 kJ .
