When 100mL of 1.00 M Ba (NO3)2 and 100 mL of 1.00 M Na2SO4, both at 25 degrees celsius, are mixed in a constant pressure calorimeter, a white precipitate is formed and the temperature rises to 28.1 degrees celsius.
A. What is the precipitate?
B. If all of the heat is absorbed by the solvent (water), what is the molar enthalpy change for the reaction.
(s for water = 4.18 J/g.C)
I really want to understand this problem so if someone could show me step by step I would really appreciate it so much!
A. Here is a simplified list of the solubility rules. You can find the material that ppts by studying this list.
http://www.files.chem.vt.edu/RVGS/ACT/notes/solubility_rules.html
B. How much heat is absorbed by the water? That is q = heat absorbed = mass H2O x specific heat water x delta T. How many moles of the ppt do you have. (hint: Write the equation an balance it to find moles ppt). Then deltaH q/mol ppt.
Post your work if you get stuck.
So for part A i balanced the equation to:
Ba(NO3)2 + Na2 -> Ba(SO4) + 2Na(NO3)
then I got: (NO3)2 + Na2 -> 2NaNO3 does that mean 2NaNO3 is the precipitate?
For part B I got q= 233.5 J
and now I'm stuck lol
A. No, it doesn't mean NaNO3 is the ppt. You didn't read the table in the link I gave you. That link tells you that sulfates are soluble EXCEPT for (see #4 in the list).
Ba(NO3)2(aq) + Na2SO4(aq) ==> BaSO4(s) + 2NaNO3(aq).
B. 233.5 is not right.I don't know what show your work means to you but it doesn't mean to show the answer only. I'm headed to bed and I don't have time to wait around while we go through this so the problem follows:
mass x specific heat x delta T
200 g H2O x 4.18 x (28.1-25) = ?? = q
delta H/mol = q/mol = q/0.1 = ??
To answer this question, we need to analyze the reaction between barium nitrate (Ba(NO3)2) and sodium sulfate (Na2SO4). Given that a white precipitate is formed, it suggests that a double displacement reaction has occurred, resulting in the formation of an insoluble compound.
A. To determine the precipitate formed, we need to examine the solubility rules for common compounds:
1. All nitrates (NO3-) are soluble.
2. All sodium salts (including Na2SO4) are soluble.
3. Most barium salts (including Ba(NO3)2) are soluble, except for barium sulfate (BaSO4) and a few others.
According to these rules, the precipitate formed in this reaction is barium sulfate (BaSO4).
B. To calculate the molar enthalpy change for the reaction, we need to use the equation:
q = m * c * ΔT
Where:
q = heat absorbed or released (in joules)
m = mass of the solvent (water) (in grams)
c = specific heat capacity of the solvent (water) (in J/g°C)
ΔT = change in temperature (in °C)
In this problem, all of the heat is absorbed by the solvent, which is water. Thus, we can set up the equation as follows:
q = m * c * ΔT
To solve for q, we need to determine the mass of water. We can use the density of water, which is 1 g/mL, to convert the volumes given into grams:
Mass of water = volume of water * density of water
Given that equal volumes (100 mL) of Ba(NO3)2 and Na2SO4 solutions are used, the total volume of the mixture is 200 mL. Thus, the mass of water can be calculated as:
Mass of water = 200 mL * 1 g/mL = 200 g
Now let's substitute the values into the equation:
q = m * c * ΔT
q = 200 g * 4.18 J/g°C * (28.1 - 25) °C
The change in temperature (ΔT) is 28.1 °C - 25 °C = 3.1 °C. Plugging in the values:
q = 200 g * 4.18 J/g°C * 3.1 °C
q = 2484.4 J
The heat absorbed (q) is calculated as 2484.4 Joules.
To find the molar enthalpy change (ΔH) for the reaction, we need to convert the heat absorbed (q) into the molar quantity of the limiting reagent. In this case, Ba(NO3)2 is the limiting reagent because it will react completely with Na2SO4, forming the BaSO4 precipitate.
To calculate the molar enthalpy change, we use the equation:
ΔH = q / n
Where:
ΔH = molar enthalpy change (in J/mol)
q = heat absorbed (in J)
n = moles of the limiting reagent.
To find the moles of Ba(NO3)2, we use the equation:
moles = Molarity * volume (in L)
Given that the volume of Ba(NO3)2 solution is 100 mL (0.100 L) and the Molarity is 1.00 M, we can calculate the moles of Ba(NO3)2 as:
moles of Ba(NO3)2 = 1.00 mol/L * 0.100 L = 0.100 mol
Since Ba(NO3)2 has a 1:1 stoichiometric ratio with BaSO4, the moles of Ba(NO3)2 equal the moles of BaSO4 precipitate formed.
Now we can calculate the molar enthalpy change:
ΔH = q / n
ΔH = 2484.4 J / 0.100 mol
ΔH = 24844 J/mol
Therefore, the molar enthalpy change (ΔH) for the reaction is 24844 J/mol.