Simple equation aluminum wire reacting with excess copper (II) nitrate

Question

15.0 grams of aluminum wire completely reacts with excess copper (II) nitrate in solution. How many grams of copper are produced?

i tried to write the equation: 2Al (s) 3Cu(NO3)2 (aq) >> 3Cu (s) Al(NO3)3 (aq)

Then i multiplied the 15 grams by 190.65 (molecular mass of 3Cu) ans then divided that by 53.96 (molecular mass of 2 Al) and my answer was 52.99. Is my answer correct or did i make a mistake? Are my phases of matter correct? Also, is this a single replacement reaction?

Answer 1

2 Al + 3Cu(NO3)2 >> 2 Al(NO3)3 + 3 Cu

Moles Al = 15.0 g / 26.98 g/mol =0.556

Moles Cu = 0.556 x 3 / 2 =0.834

mass Cu = 0.834 mol x 63.546 g/mol =53.0 g