If 4.42 grams of iron reacts how much iron(III)oxide is produced

i need an answers i got 6.3218 grams of iron (III) oxide i need to check my answer is this a right answer

4Fe + 3O2 -----> 2Fe2O3

Molecular Mass of Fe2O3 = 159.7g/mol

Atomic Mass of Fe = 55.85g/mol

4.42g / 58.85g/ mol = 0.079moles of Fe

In our balanced equation the ratio between Fe and Fe2O3 is 2:1 therefore

(moles of Fe) / 2 = moles of Fe2O3.

(0.079moles / 2) * 159.7g/mol = 6.32 grams of Fe2O3

4Fe + 3O2 --> 2Fe2O3

4.42g Fe / 55.8g/mole = 0.079moles Fe
0.079moles Fe x (2Fe2O3 / 4Fe) x 159.6g/mole = 6.3g Fe2O3 produced