What mass of propane (c3h8(g)) must be burned to supply 2775 kj of heat? the standard enthalpy of combustion of propane at 298 k is −2220 kj · mol−1 ?

Answer is: 55.125 grams of propane must be burned.
Balanced chemical reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
Make proportion: 1 mol(C₃H₈) : 2220 kJ = n(C₃H₈) : 2775kJ.
n(C₃H₈) = 2775 kJ·mol ÷ 2220 kJ.
n(C₃H₈) = 1.25 mol.
m(C₃H₈) = n(C₃H₈) · M(C₃H₈).
m(C₃H₈) = 1.25 mol · 44.1 g/mol.
m(C₃H₈) = 55.125 g.
n - amount of substance.

The answer is 4 i don't know if its right let me see.