Use the following information to identify element A and come
pound B, then answer questions a and b.
An empty glass container has a mass of 658.572 g. It has a
mass of 659.452 g after it has been filled with nitrogen gas at a
pressure of 790. torr and a temperature of 15°C. When the container
is evacuated and refilled with a certain element (A) at a
pressure of 745 torr and a temperature of 26"C, it has a mass of
660.59 g.
Compound B, a gaseous organic compound that consists of
85.6% carbon and 14.4% hydrogen by mass, is placed in a stainless
steel vessel (10.68 L) with excess oxygen gas. The vessel is
placed in a constant-temperature bath at 22°C. The pressure in
the vessel is 1 1.98 atm. In the bottom of the vessel is a container
that is packed with Ascarite and a desiccant. Ascarite is asbestos
impregnated with sodium hydroxide; it quantitatively absorbs
carbon dioxide:
2NaOH(s) + CO2(g) -----> Na2CO3(s) H2O(l)
The desiccant is anhydrous magnesium perchlorate, which
quantitatively absorbs the water produced by the combustion reaction
as well as the water produced by the above reaction. Neither
the Ascarite nor the desiccant reacts with compound B or
oxygen. The total mass of the container with the Ascarite and
desiccant is 765.3 g.
The combustion reaction of compound B is initiated by a
spark. The pressure immediately rises, then begins to decrease,
and finally reaches a steady value of 6.02 atm. The stainless steel
vessel is carefdly opened, and the mass of the container inside
the vessel is found to be 846.7 g.
A and B react quantitatively in a 1: 1 mole ratio to form one
mole of the single product, gas C.
a. How many grams of C will be produced if 10.0 L of A and
8.60 L of B (each at STP) are reacted by opening a stopcock
connecting the two samples?
b. What will be the total pressure in the system?
I have worked through part of the problem; at least enough to know how to finish.
Use the first paragraph to determine the identity of element A.
The difference in mass of the empty container and the N2 filled container will allow you to calculate the volume of the container. Then the filling with element A gas allows you to calculate the number of mols of A and that, along with the mass of element A, will give you the molar mass of element A.
Compound B give %C and %H which allows you to calculate the empirical formula for B.
When you get this far, post with your work if you don't know how to continue and we can go from there.
Okay I am working on this problem and I found A to be Cu and compound B to be CH2. Is this correct? And I am totally lost on the rest of the problem. Could you please just tell me where to start?
Great job on identifying element A as copper (Cu) and compound B as CH2. Now let's move on to the rest of the problem.
a. To find the grams of C produced when 10.0 L of A and 8.60 L of B (both at STP) are reacted, we need to find the limiting reactant first. From the balanced equation, we know that 1 mole of A reacts with 1 mole of B to produce 1 mole of C.
First, we need to calculate the number of moles of A and B:
- For A: Since we are given that 10.0 L of A is at STP, we can use the ideal gas law (PV = nRT) to calculate the number of moles of A. At STP (standard temperature and pressure), 1 mole of gas occupies 22.414 L, so we have:
nA = PV/RT = (1 atm)(10.0 L)/(0.0821 atm·L/mol·K)(273 K) = 0.4605 mol
- For B: Similarly, we have 8.60 L of B at STP, so:
nB = PV/RT = (1 atm)(8.60 L)/(0.0821 atm·L/mol·K)(273 K) = 0.3982 mol
Since the molar ratio of A to B is 1:1, we can see that B is the limiting reactant because it has fewer moles. Therefore, we will use the number of moles of B to calculate the moles of C produced.
- For C: Since 1 mole of B reacts to produce 1 mole of C, we have:
nC = nB = 0.3982 mol
Next, we need to calculate the molar mass of compound B to find the grams of C produced. The empirical formula of CH2 tells us the molar mass of B is 14.01 g/mol + (1.008 g/mol x 2) = 16.02 g/mol.
Now, using the equation grams = moles x molar mass, we can calculate the grams of C:
grams of C = nC x molar mass of C = 0.3982 mol x 16.02 g/mol = 6.366 g
Therefore, 6.366 grams of C will be produced when 10.0 L of A and 8.60 L of B are reacted.
b. To find the total pressure in the system, we need to consider the partial pressures of the gases in the system. Since we are given the pressure of the combustion reaction is 6.02 atm, we can assume that's the pressure contributed by the produced gas C.
Since we found that 0.3982 mol of B produces 0.3982 mol of C, and the volumes are the same at STP, we can use the ideal gas law to find the partial pressure of C:
PV = nRT
(6.02 atm)(V) = (0.3982 mol)(0.0821 atm·L/mol·K)(273 K)
V = (0.3982 mol)(0.0821 atm·L/mol·K)(273 K) / (6.02 atm)
V ≈ 4.2 L
Therefore, the partial pressure of gas C is 6.02 atm, and the total pressure in the system would be the sum of the partial pressure of C and the initial pressure of B, which is 11.98 atm. So the total pressure in the system would be 6.02 atm + 11.98 atm = 18.0 atm.
In summary:
a. 10.0 L of A and 8.60 L of B (at STP) will produce 6.366 grams of C.
b. The total pressure in the system will be 18.0 atm.