What mass of a 5% solution of copper (II) nitrate will react with 250 g of a 10% solution of sodium hydroxide?

Given:
w (Cu (NO3) 2) = 5% = 0.05
m p (NaOH) = 250 g
w (NaOH) = 10% = 0.1
To find:
m p (Cu (NO3) 2)
Decision:
Cu (NO3) 2 + 2NaOH = 2NaNO3 + Cu (OH) 2
m in (NaOH) = m p * w = 250 g * 0.1 = 25 g
n (NaOH) = m / M = 25 g / 40 g / mol = 0.625 mol
n (NaOH): n (Cu (NO3) 2) = 2: 1
n (Cu (NO3) 2) = 0.625 mol / 2 = 0.313 mol
m in (Cu (NO3) 2) = n * M = 0.313 mol * 188 g / mol = 59 g
m p (Cu (NO3) 2) = m w / w = 59 g / 0.05 = 1180 g
Answer: 1180 g