When 4.76 g of a solid mixture composed of NH4Cl and CaCl2 was dissolved in 100.0 mL of water, the temperature of the water rose by 4.73°C. How many grams of each substance was in the mixture?
Madapaka
To find the mass of each substance in the mixture, we can use the concept of heat transfer.
First, we need to calculate the heat transferred to the water:
q = mcΔT
Where:
q = heat transferred
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature
Given:
m = 100.0 mL of water = 100.0 g (since the density of water is approximately 1 g/mL)
c = 4.18 J/g°C (specific heat capacity of water)
ΔT = 4.73°C
Substituting the values into the equation:
q = (100.0 g) x (4.18 J/g°C) x (4.73°C)
q ≈ 1967.9 J
Next, we need to calculate the moles of NH4Cl and CaCl2 in the mixture.
moles of substance = mass / molar mass
Molar mass of NH4Cl = 53.49 g/mol
Molar mass of CaCl2 = 110.98 g/mol
Let's assume the mass of NH4Cl is "x" grams and the mass of CaCl2 is "y" grams:
x + y = 4.76 g (Equation 1)
The moles of NH4Cl can be calculated as:
moles of NH4Cl = x / 53.49 g/mol (Equation 2)
The moles of CaCl2 can be calculated as:
moles of CaCl2 = y / 110.98 g/mol (Equation 3)
Now, we can set up an equation based on the heat of dissolution and the number of moles:
q = (moles of NH4Cl x heat of dissolution of NH4Cl) + (moles of CaCl2 x heat of dissolution of CaCl2)
The heat of dissolution for NH4Cl is 14.8 J/g, and for CaCl2 is 81.9 J/g.
Substituting the values and equations:
1967.9 J = [(x / 53.49 g/mol) x 14.8 J/g] + [(y / 110.98 g/mol) x 81.9 J/g]
Now we can solve the system of equations (Equation 1 and the above equation) to find the values of 'x' and 'y'.
To solve this problem, we will use the principles of calorimetry. Calorimetry is the science of measuring heat transfer, and it allows us to determine the amount of heat gained or lost by a substance. In this case, we can use the equation:
q = m * Cp * ΔT
Where:
- q is the heat transferred or the energy change (in Joules),
- m is the mass of the substance (in grams) that is undergoing the temperature change,
- Cp is the specific heat capacity of the substance (in J/g°C), and
- ΔT is the change in temperature (in °C).
First, we need to calculate the heat transfer (q) when the solid mixture is dissolved in the water. In this case, the heat transferred to the water is equal to the heat absorbed by the solid mixture. The specific heat capacity of water (Cp) is approximately 4.18 J/g°C.
Given:
- mass of the solid mixture (m) = 4.76 g
- change in temperature (ΔT) = 4.73°C
- specific heat capacity of water (Cp) = 4.18 J/g°C
Using the equation q = m * Cp * ΔT, we can calculate the heat transfer (q) to the water.
q = 4.76 g * 4.18 J/g°C * 4.73°C
q ≈ 95.24 J
The next step is to determine the amount of NH4Cl and CaCl2 in the mixture. Let's assume that x grams of NH4Cl and y grams of CaCl2 are present.
NH4Cl: molar mass = 53.49 g/mol
CaCl2: molar mass = 110.98 g/mol
We can calculate the moles of NH4Cl (n1) and moles of CaCl2 (n2) using their respective molar masses:
n1 = x / 53.49 mol
n2 = y / 110.98 mol
Since there is no change in moles during the dissolution process, we have the following relationship:
n1 + n2 = n
Here, n is the total number of moles, and it is equal to the total mass of the solid mixture divided by its molar mass:
n = 4.76 g / (x/53.49 + y/110.98) mol
Finally, we can consider the heat transfer as the sum of the heat transferred by NH4Cl and CaCl2. The heat transfer (q) can be expressed in terms of moles and their respective enthalpies of dissolution (ΔH), as follows:
q = n1 * ΔH1 + n2 * ΔH2
The enthalpies of dissolution for NH4Cl and CaCl2 are exothermic processes, so their values are negative. For NH4Cl, ΔH1 = -15.23 kJ/mol, and for CaCl2, ΔH2 = -81.3 kJ/mol.
Converting ΔH1 and ΔH2 to J/mol, we have:
ΔH1 = -15.23 kJ/mol * 1000 J/kJ = -15230 J/mol
ΔH2 = -81.3 kJ/mol * 1000 J/kJ = -81300 J/mol
Therefore, we can write the equation for q:
q = (x / 53.49) * (-15230 J/mol) + (y / 110.98) * (-81300 J/mol)
Now, using the equation q = 95.24 J, we can solve for x and y to find the grams of NH4Cl and CaCl2 in the mixture. However, the calculations involving changing one equation into two linear equations are quite complex, so it will take some time to solve them.
This will take some work but I would look at setting up two simultaneous equations and solving them.
You know the heat is q = mass x specific heat x delta T ort
q = 100 x 4.184 x 4.73 = ? J. Change to kJ.
let X = grams NH4Cl
and Y = grams CaCl2. so
X + Y = 4.76 is the first equation.
Look up the heat of dissolution for CaCl2 and NH4Cl and set up the second equation from that in terms of X and Y. I think NH4Cl is about +15 kJ/mol(endothermic) and CaCl2 is about -80 kJ/mol but you need to look them up and use the correct value. .How many mols NH4Cl do you have? That's X/mm NH4Cl where mm is molar mass. mols CaCl2 is Y/mm CaCl2. So
heat solution NH4Cl*(X/mmNH4Cl) + heat sol CaCl2*(Y/mmCaCl2) = q in kJ/mol
Solve for X and Y in grams NH4Cl and CaCl2.
If you get stuck post ALL of your work.