Steam at 100°C was passed into a flask containing 250 g of water at 21°C, where the steam condensed. How many grams of steam must have condensed if the temperature of the water in the flask was raised to 76°C? The heat of vaporization of water at 100°C is 40.7 kJ/mol and the specific heat is 4.18 J/(g·°C).

I use the formula
q=m*specific heat*delta T
250*4.18*(76-21) = 57.475KJ
Hvap/g H20
40.7/18 = 2.26KJ/g
q=Hvap/gh20
57.475KJ/2.26KJ/g = 25.4g (answer)

It keeps telling me the answer is wrong. Can you please tell me what I may be doing wrong in this problem?

Thank you for showing your work. I think you have failed to take into account the added water from the steam AND that the final T is NOT 100 C.

So I think you have three items at play here. Steam condenses at 100 to liquid water (which adds to the mass of the water) and that heat is Hvap x mass steam. Mass steam is the unknown here.
Then you have the mass of the water from the steam x sp.h. x delta T (which is Tfinal-Tinitial = 76-100). Finally you have the original 250 g H2O being heated up from 21 to 76 (or Tf-Ti = +55). The + signifies that heat is being added. The steam condenses to give off heat so that portion of the equation is negative and the 100 water then cools to 76 (Tf-Ti=76-100=-24) so it is negative also. I hope this helps.

Your calculation steps and formulas are correct, but there is a mistake in the conversion factor you used to convert the heat of vaporization from kJ/mol to kJ/g.

The correct conversion factor should be:
40.7 kJ/mol / (18 g/mol) = 2.261 kJ/g (rounded to three decimal places)

Using this correct conversion factor in the final calculation:
57.475 kJ / 2.261 kJ/g = 25.422 g (rounded to three decimal places)

The correct answer is approximately 25.422 grams, not 25.4 grams.

Based on the information provided, your approach and calculations appear to be correct. However, there might be a minor mistake in your final step.

Let's double-check the calculation:

First, calculate the heat absorbed by the water using the formula:
q = m * specific heat * delta T
q = 250 g * 4.18 J/(g·°C) * (76°C - 21°C)
q = 78,685 J

Next, convert the heat absorbed from joules to kilojoules (since the heat of vaporization is given in kJ/mol):
q = 78,685 J / (1000 J/kJ)
q = 78.685 kJ

Now, divide the heat absorbed by the heat of vaporization per gram of water:
grams of steam condensed = q / heat of vaporization per gram of water
grams of steam condensed = 78.685 kJ / 2.26 kJ/g
grams of steam condensed ≈ 34.8 g (rounded to one decimal place)

Therefore, according to the calculations, approximately 34.8 grams of steam must have condensed.

Make sure to recheck your calculations and units to identify any errors that might lead to the incorrect answer.