Please help me with this question:

Calculate the theoretical yield amount of CaCl2 that would be required to change the temp of 50ml of water from room temp 20degree to 45 degrees. Heat Capacity H20 is 4.184 Jg^-1deg^-1 and density is 1g/mL.

I've already calculated: Q=mc delta T
(50g)(4.184J/g^-1)(45-20deg)
=5230 J

so then the Q of CaCl2 is -5230 J. What do I do from here I am so confused!

You need to know the grams (or moles) of CaCl2 needed to produce a change of x joules.

To calculate the theoretical yield amount of CaCl2 required to achieve the given temperature change, you need to consider the heat exchange that occurs between the CaCl2 and the water. Here's how you can proceed:

Step 1: Determine the molar mass of CaCl2.
The molar mass of CaCl2 is calculated by adding the atomic masses of calcium (Ca) and two chlorine atoms (Cl). The atomic mass of Ca is 40.08 g/mol, and the atomic mass of Cl is 35.45 g/mol. So the molar mass of CaCl2 is:
Molar mass = (1 * Ca) + (2 * Cl) = 40.08 g/mol + (2 * 35.45 g/mol) = 110.98 g/mol.

Step 2: Calculate the amount of heat (q) from the water using the equation you already determined:
q = mcΔT = (50 g) * (4.184 J/g·°C) * (45 - 20) °C = 5230 J.

Step 3: Convert the heat (q) to moles of CaCl2.
Since q is the heat exchanged by the water, it is also the heat absorbed by the CaCl2. To convert the heat to moles, we need to use the molar heat capacity (Cp) of CaCl2, which is the amount of heat required to raise one mole of CaCl2 by one degree Celsius. The molar heat capacity of CaCl2 is typically given as 109 J/mol·°C.

To convert q to moles, we use the following equation:
q = nCpΔT

Since we are solving for n (the number of moles), we rearrange the equation as follows:
n = q / (CpΔT)

Substituting the known values:
n = 5230 J / (109 J/mol·°C * 1 °C) ≈ 47.98 mol

Step 4: Convert moles to grams using the molar mass.
To convert moles of CaCl2 to grams, you multiply the number of moles by the molar mass:
Mass of CaCl2 = n * Molar mass = 47.98 mol * 110.98 g/mol ≈ 5329.26 g.

So, the theoretical yield amount of CaCl2 required to achieve the given temperature change in 50 mL of water is approximately 5329.26 grams.