Answer:
About 67 grams or 67.39 grams
Explanation:
First you would have to remember a few things:
enthalpy to melt ice is called enthalpy of fusion. this value is 6.02kJ/mol
of ice
it takes 4.18 joules to raise 1 gram of liquid water 1 degree C
water boils at 100 degrees C and water melts above 0 degrees C
1 kilojoules is 1000 joules
water's enthalpy of vaporization (steam) is 40.68 kJ/mol
a mole of water is 18.02 grams
we also have to assume the ice is at 0 degrees C
Step 1
Now start with your ice. The enthalpy of fusion for ice is calculated with this formula:
q = n x ΔH q= energy, n = moles of water, ΔH=enthalpy of fusion
Calculate how many moles of ice you have:
150g x (1 mol / 18.02 g) = 8.32 moles
Put that into the equation:
q = 8.32 mol x 6.02 = 50.09 kJ of energy to melt 150g of ice
Step 2
To raise 1 gram of water to the boiling point, it would take 4.18 joules times 100 (degrees C) or 418 joules.
So if it takes 418 joules for just 1 gram of water, it would take 150 times that amount to raise 150g to 100 degrees C. 418 x 150 = 62,700 joules or 62.7 kilojoules.
So far you have already used 50.09 kJ to melt the ice and another 62.7 kJ to bring the water to boiling. That's a total of 112.79 kJ.
Step 3
The final step is to see how much energy is left to vaporize the water.
Subtract the energy you used so far from what you were told you have.
265 kJ - 112.79 kJ = 152.21 kJ
Again q = mol x ΔH (vaporization)
You know you only have 152.21 kJ left so find out how many moles that will vaporize.
152.21 kJ = mol x 40.68 or mol = 152.21 / 40.68 = 3.74 moles
This tells you that you have vaporized 3.74 moles with the energy you have left.
Convert that back to grams.
3.74 mol x ( 18.02 g / 1 mol ) = 67.39 grams
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