A quantity of ice at 0.0 °C was added to 33.6 g of water at 45.0 °C to give water at 0.0 °C. How much ice was added? The heat of fusion of water is 6.01 kJ/mol, and the specific heat is 4.18 J/(g•°C).

Answer 1

The ice at zero is to melt and remain at zero. So the heat fusion must be enough to lower the liquid water from 45.0 to zero C. How much heat must be removed from the water. That's
q = mass water x specific heat H2O x (Tfinal-Tinitial). That gives you q in J.

Now convert heat fusion from kJ/mol to J/g. That's approx 334 J/g but you need to look up the exact number.
Then how much ice is needed to furnish that q.
q for melting is mass ice x heat fusion. Solve for mass ice to produce that q.