How much heat is required to convert 9.00 g of water at 100°C into steam at 100°C, if ∆Hvap for water = 40.7 kJ/mol?
To calculate the amount of heat required to convert 9.00 g of water at 100°C to steam at 100°C, we need to follow these steps:
Step 1: Convert mass to moles
Use the molar mass of water (H2O) to convert the mass of water to moles.
Molar mass of H2O = 2g/mol (2 hydrogen atoms) + 16g/mol (1 oxygen atom) = 18 g/mol
Number of moles = mass / molar mass
Number of moles = 9.00 g / 18 g/mol = 0.50 mol
Step 2: Calculate the heat required
The enthalpy of vaporization (∆Hvap) is given as 40.7 kJ/mol. Since we have 0.50 mol of water, we can calculate the heat required using the following formula:
Heat required = ∆Hvap x number of moles
Heat required = 40.7 kJ/mol x 0.50 mol = 20.35 kJ
Therefore, the amount of heat required to convert 9.00 g of water at 100°C into steam at 100°C is 20.35 kJ.
To calculate the amount of heat required to convert water at 100°C into steam at 100°C, you'll need to follow a series of steps.
1. Determine the number of moles of water:
- We are given the mass of water, which is 9.00 g.
- To convert grams to moles, we need to use the molar mass of water, which is approximately 18.015 g/mol.
- Divide the mass of water by the molar mass to get the number of moles. In this case, it is 9.00 g / 18.015 g/mol = 0.499 mol.
2. Calculate the heat required using the given value for ∆Hvap:
- ∆Hvap is the heat of vaporization, which is the amount of energy required to convert one mole of a substance from a liquid to a gas at its boiling point. In this case, ∆Hvap for water is given as 40.7 kJ/mol.
- Multiply the moles of water by the ∆Hvap to get the total heat required. In this case, it is 0.499 mol * (40.7 kJ/mol) = 20.2843 kJ.
So, the amount of heat required to convert 9.00 g of water at 100°C into steam at 100°C is approximately 20.28 kJ.
Convert 9 g H2O to mols. mols = grams/molar mass = ? Then
q = dH(in kJ/mol) x mols H2O = ?