;-; pls help me?

1)
Molar mass of C₂H₄ = (12.0×2 + 1.0×4) g/mol = 28.0 g/mol
Moles of C₂H₄ reacted = (4.79 g) / (28.0 g/mol) = 0.171 mol

C₂H₄(g) + 3O₂(g) → CO₂(g) + 2H₂O(l) …… ΔH = -1.39 × 10³ kJ
The reaction of 1 mol C₂H₄ liberates 1.39 × 10³ kJ of heat.
Heat liberated = (0.171 mol) × (1.39 × 10³ kJ) = 238 kJ (to 3 sig. fig.)

OR:
(4.79 g C₂H₄) × (1 mol C₂H₄ / 28.0 g/mol) × (1.39 × 10³ kJ of energy / 1 mol C₂H₄)
= 238 kJ (to 3 sig. fig.)

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2)
C(s) + 2S(s) → CS₂(l) …… ΔH = 89.3 kJ
89.3 kJ of energy are available when 1 mol of C reacts.
Moles of C burned = (89.3 kJ) / (520.0 kJ/mol) = 0.172 mol

Molar mass of C = 12.0 g/mol
Mass of C reacted = (0.172 mol) × (12.0 g/mol) = 2.06 g (to 3 sig. fig.)

OR:
(89.3 kJ heat) × (1 mol C / 520.0 kJ) × (12.0 g C / 1 mol C) = 2.06 g (tp 3 sig. fig.)

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3)
Energy required to change 40.0 g of liquid water from 23.0°C to 100.0°C
= m c ΔT
= (40.0 g) × (4.186 J/g°C) × (100.0 - 23.0)°C
= 12890 J
= 12.9 kJ

Energy required to change 40.0 g of liquid water to steam at 100.0°C
= m Lv
= (40.0 g) × (2260 J/g)
= 90400 J
= 90.4 kJ

Energy required to change 40.0 g of steam from 100.0°C to 145.0°C
= m c ΔT
= (40.0 g) × (1.996 J/g°C) × (145.0 - 100.0°C)
= 360 J
= 3.6 kJ

Energy required to change 40.0 g of liquid water at 23°C to steam at 145.0°C
= (12. 9 + 90.4 + 3.6) kJ
= 107 kJ (to 3 sig. fig.)

(The answer would be slightly different when different sources of values of c and Lv are used.)