How many grams of CaO must be added to 50.0 mL of water to make the temperature increase from 25.0 ¡ÆC to 48.81 ¡ÆC? The reaction is

CaO(s) + H2O(l) ¡ê Ca(OH)2(aq) -- ¥ÄH = -83.7 kJ

Assume the heat capacity of the solution is the same as pure water (4.184 J/g*¡ÆC), the density of the solution is 1.00 g/mL and there is no loss of heat to the

To find the amount of CaO needed to increase the temperature of water from 25.0 °C to 48.81 °C, we need to calculate the heat gained by the water and equate it to the heat released by the CaO.

Step 1: Calculate the heat gained by the water.
The heat formula is: q = m * c * ΔT, where q is the heat gained/lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Given:
- Volume of water = 50.0 mL
- Density of the solution = 1.00 g/mL
- Specific heat capacity of water (c) = 4.184 J/g*°C
- Initial temperature (T1) = 25.0 °C
- Final temperature (T2) = 48.81 °C

First, we calculate the mass of the water:
mass = volume * density
mass = 50.0 mL * 1.00 g/mL
mass = 50.0 g

Now, calculate the heat gained by the water:
q = m * c * ΔT
q = 50.0 g * 4.184 J/g*°C * (48.81 °C - 25.0 °C)

Step 2: Calculate the heat released by CaO.
Given:
- Enthalpy change of the reaction (ΔH) = -83.7 kJ

Convert the ΔH value to joules:
ΔH = -83.7 kJ * 1000 J/kJ
ΔH = -83,700 J

Step 3: Set up the equation and solve for the mass of CaO.
Since the heat gained by the water is equal to the heat released by CaO (assuming no heat loss), we can equate the two values and solve for the mass of CaO:
q = -ΔH

50.0 g * 4.184 J/g*°C * (48.81 °C - 25.0 °C) = -83,700 J

Solving for the mass of CaO:
mass_CaO = (-83,700 J) / (50.0 g * 4.184 J/g*°C * (48.81 °C - 25.0 °C))

The resulting mass of CaO will be in grams, and that is the amount needed to increase the temperature of the water.

To solve this problem, we need to use the equation:

q = (m * C * ΔT) + (m * ΔH)

where:
q is the heat absorbed by the solution
m is the mass of the solution
C is the heat capacity of the solution
ΔT is the change in temperature
ΔH is the enthalpy change of the reaction

We are looking for the mass of CaO, so let's calculate the values one by one:

1. Calculate the heat absorbed by the solution (q):
q = (m * C * ΔT) + (m * ΔH)

Given:
C = 4.184 J/g*°C (heat capacity of the solution)
ΔT = (48.81 - 25.0) °C = 23.81 °C (change in temperature)
ΔH = -83.7 kJ = -83700 J (enthalpy change of the reaction)

Substituting the values, we have:
q = (m * 4.184 J/g*°C * 23.81°C) + (m * -83700 J)

2. Convert mL to grams (since density is given):
1 mL = 1 g

Given:
V = 50.0 mL (volume of water)
density = 1.00 g/mL

Substituting the values, we have:
m = V * density
m = 50.0 mL * 1.00 g/mL
m = 50.0 g

3. Substitute the values of q and m into the equation:
q = (m * 4.184 J/g*°C * 23.81°C) + (m * -83700 J)

4. Solve for m (mass of CaO):
We know that CaO reacts with H2O in a 1:1 ratio. This means that the moles of CaO reacting is equal to moles of Ca(OH)2 formed.

Given:
Molar mass of CaO = 56.08 g/mol
Molar mass of Ca(OH)2 = 74.09 g/mol

The molar ratios are:
1 mol CaO reacts to form 1 mol Ca(OH)2

5. Convert mass of Ca(OH)2 to moles:
moles of Ca(OH)2 = mass / molar mass
moles of Ca(OH)2 = q / (molar mass * ΔH)

6. Convert moles of Ca(OH)2 to mass of CaO:
mass of CaO = moles of Ca(OH)2 * molar mass of CaO

Now we can solve the problem step-by-step.