What mass of a 15% copper (II) nitrate solution will react with 120 g of a 10% potassium hydroxide solution?

Given:
w (Cu (NO3) 2) = 15% = 0.15
m p. (KOH) = 120 g
w (KOH) = 10% = 0.1
To find:
m p. (Cu (NO3) 2
Decision:
Cu (NO3) 2 + 2KOH = Cu (OH) 2 ↓ + 2KNO3
m in. (KOH) = w * m p. = 0.1 * 120 g = 12 g
n (KOH) = m: M = 12 g: 56 g / mol = 0.214 mol
n (KOH): n (Cu (OH) 2) = 2: 1
n (Cu (OH) 2) = 0.214 mol: 2 = 0.107 mol
n (Cu (OH) 2): n (Cu (NO3) 2 = 1: 1
n (Cu (NO3) 2 = 0.107 mol
m in. (Cu (NO3) 2 = n * M = 0.107 mol * 188 g / mol = 20.12 g
m p. (Cu (NO3) 2 = m in.: w = 20.12 g: 0.15 = 134 g
Answer: m p. (Cu (NO3) 2) = 134 g