[ahawk1]

You drop 225 g of ice into 500.0 mL of lukewarm water in an insulated pitcher. The water is
initially at 25.0°C and the ice cubes are initially at –6.0°C. When the contents of the pitcher reach
thermal equilibrium, both ice and water are present at 0.0°C. How many grams of ice will still be
present?
The density of water is 1.0 g/mL, and the heat of fusion for water is 6.01 kJ/mol. The heat capacity
of liquid water is 75.3 J/K/mol, and the heat capacity of ice is 37.7 J/K/mol.

this is what i did:
-qwater=qice
-75.3*25*500=37.7*6*225+6011*n
what is wrong with this? thanks!

[Astrokel]

If not all ice are melted, why do you multiply it by 225?

[ahawk1]

If not all ice are melted, why do you multiply it by 225?

so would i do this:
-75.3*25*500=37.7*6*m+6011*(m/18)

[Astrokel]

You can drop the minus sign. What you are doing are is heat lost = heat gained. So assume m grams of ice has been melted, your equation is right just take note the specific heat is in unit mol. After getting your value m, you can calculate the mass of ice has not melted.

[ahawk1]

You can drop the minus sign. What you are doing are is heat lost = heat gained. So assume m grams of ice has been melted, your equation is right just take note the specific heat is in unit mol. After getting your value m, you can calculate the mass of ice has not melted.

thanks for everything!!!