Percent yield question! chem check my answer please?

Molar mass of C₇H₈ = (12.0×7 + 1.0×8) g/mol = 92.0 g/mol
Molar mass of HNO₃ = (1.0 + 14.0 + 16.0×3) g/mol = 63.0 g/mol
Molar mass of C₇H₅(NO₂)₃ = (12.0×7 + 1.0×5 + 14.0×3 + 16.0×6) g/mol = 213.0 g/mol

C₇H₈ + 3HNO₃ → C₇H₅(NO₂)₃ + 3H₂O
OR: Mole ratio C₇H₈ : HNO₃ = 1 : 3

Initial no. of moles of C₇H₈ = (454 g) / (92.0 g/mol) = 4.93 mol
Initial no. of moles of HNO₃ = (1000 g) / (63.0 g/mol) = 15.9 mol

If C₇H₈ completely reacts, HNO₃ needed = (4.93 mol) × 3 = 14.8 mol < 15.9 mol
Hence, HNO₃ is in excess, and thus C₇H₈ completely reacts (the limiting reactant).

According to the above equation, mole ratio C₇H₈ : C₇H₅(NO₂)₃ = 1 : 1
No. of moles of C₇H₈ (the limiting reactant) reacted = 4.93 mol
Theoretically, no. of moles of C₇H₅(NO₂)₃ produced = 4.93 mol
Theoretical yield of C₇H₅(NO₂)₃ = (4.93 mol) × (213.0 g/mol) = 1050 g
% yield = (1050/1080) × 100% = 97.2%