How many grams of ice are needed to cool 250 ml of water from 75 degrees Celsius to 5 degrees Celsius?

Question

(m_ice * 6.01kJ) + (m_ice * 4.18J/g°C * (5-75°C) = -[250mL * 4.18J/g°C * (5-75°C)] ???

Answer 1

Most is ok but that second term I think should be mice*4.18*(5-0). At any rate, I would change the 6.01 kJ/mol to J/g*C = 334 J/g*C
Then [mice*334] + [mice*4.18*(5-0)] + [250 x 4.18 x (5-75) = 0
I solved and got close to 206 g ice needed.