I posted this question on Thursday April 12 and received some help..but need more.. thanks
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 already have A.
I have the following equation
2CH2 + 3O2 ==> 2CO2 + 2H2O
I know I must find the percentage of CO2 and H2O in a 1:1 ratio, but I want to know if I am doing the percentage correct before I move one to anything else.
1 mol CO2 will have a mass of 44.01 g.
1 mol H2O will have a mass of 18.016 g.
I did 846.7-765.3= 81.4g to get the gain in mass which is the weight of CO2 & H20.
Then for CO2 44g/81.4 *100 = 54.066%
H20:18.016/81.4g *100 =13%
But shouldn't the percentages somehow add up to 100?
The pressure of compound B + pressure of oxygen before spark ignition is 11.98 atm. After ignition, pressure is 6.02. Therefore, 11.98-6.02= 5.96 is how much reacted (due to pressure). Use PV = nRT to calculate the number of mols at the appropriate T and V. Keep several numbers (i.e., don't try to round too severely--wait until the end of the problem). This n=number of mols of B + oxygen. You will need this number later.
Next, calculate the grams gain in mass for the ascarite and magnesium perchlorate. I found 81.4 grams. How do we split that up to see how much of it is due to CO2 and how much is due to H2O? One way to do it is to look at the products of the balanced equation. Remember that was
2CH2 + 3O2 ==> 2CO2 + 2H2O
We have H2O and CO2 in 1:1 ratio.
So assume we had 2 mols CO2 and 2 mols H2O, we would have 88 mols CO2 and 36 mols H2O for a total of 124 grams. What percent is 88/124 and 36/124? Apply those percentages to the 81.4 gram gain of and that will give you grams CO2 and grams H2O. Check to make sure they add to 81.4 g.
You want to do three things with these g CO2 and g H2O.
a). Calculate g carbon in the orginal sample. Calculate g H in the original sample. Add C and H to give you the mass of the original sample, B.
b)Calculate g oxygen used to make the CO2 and calculate g oxygen used to make the H2O. Add them together, convert to mols. That will be the mols oxygen needed to convert all the C to CO2 and all the H to H2O. Now subtract mols O2 from the mols you found above for mols B + mols O2. Obviously,
mols B + mols oxygen = ??
mols oxygen = ??
subtracting gives mols B.
c)Now you have grams B and you have mols of B. Convert that to grams/mol which is the molar mass B and that will identify compound B. You already know the empirical formula is CH2 from work you did last week.
Post your work if you get stuck.
Thank you..I will try the rest!
1 mol CO2 will have a mass of 44.01 g.
1 mol H2O will have a mass of 18.016 g.
I did 846.7-765.3= 81.4g to get the gain in mass which is the weight of CO2 & H20.
OK to here.
Then for CO2 44g/81.4 *100 = 54.066%
H20:18.016/81.4g *100 =13%
But shouldn't the percentages somehow add up to 100?
Yes, the percentages should add to 100 and they will if you do them right.
44/81.4 is taking the mols ratio. That isn't what you want.
1 mol CO2 = 44.01 g
1 mol H2O = 18.016 g
Total = 62.026 g
fraction CO2 = 44.01 g/62.026 g = 0.7095, THEN 0.7095 x 81.4 = 57.757 g CO2. etc.
So I have:
n=pv/rt
n=(5.96)(10.68)/(0.082057*295) = 2.6295 moles B+ O
______________________________
70.95% CO2 =57.75g
29.05% H2O = 23.65g
_____ ________
100% 81.4g
a)
(57.75/44.01)*(1molC/1molCO2)(12.01gC)=15.76g C
(23.65g H2O/18.016g H2O)(2H/1H2O)(1.008gH)= 2.65 g H
18.4g original B
b)
(57.73g CO2/44.01g)(2O/1C)(16gO)= 41.98g O
(23.65H2O/18.016)(1/1)*16gO0=21.00gO
21.00+41.98=62.98g O2
62.98gO2/32 = 1.97moles O2
2.6295moles-1.97 = 0.6595 moles B
molar mass B: 27.9 g/mol
CH2=14.026g/mol
27.9/14.026 = 2
B= C2H4
is all that right?
You have it A-OK.
So C2H4 is what?
You now know element A and compound B, now you can do the remainder of the problem.
Yes, everything you calculated is correct. Element A is nitrogen gas (N2) and compound B is ethylene (C2H4). Now, you can move on to answering questions a and b.
a. To calculate how many grams of gas C will be produced, you need to determine the mole ratio between element A (N2) and compound B (C2H4) from the equation: 2CH2 + 3O2 → 2CO2 + 2H2O.
The mole ratio between N2 and C2H4 is 1:1, so if you have 10.0 L of A at STP (standard temperature and pressure) and 8.60 L of B at STP, you can calculate the moles of A and B using the ideal gas law equation: PV = nRT.
For element A:
P = 1 atm (since it's at STP)
V = 10.0 L
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = 273.15 K (standard temperature)
n(A) = (P × V) / (R × T)
n(A) = (1 atm × 10.0 L) / (0.0821 L·atm/(mol·K) × 273.15 K) = 0.48094 mol
For compound B:
P = 1 atm
V = 8.60 L
R = 0.0821 L·atm/(mol·K)
T = 273.15 K
n(B) = (P × V) / (R × T)
n(B) = (1 atm × 8.60 L) / (0.0821 L·atm/(mol·K) × 273.15 K) = 0.41888 mol
Since the mole ratio between A and B is 1:1, the moles of gas C produced will be the same as the moles of element A or compound B in this case.
So, the number of grams of C produced will be:
m(C) = n(C) × molar mass(C)
m(C) = 0.48094 mol × molar mass(C)
b. To find the total pressure in the system, you need to consider the partial pressures of element A (nitrogen gas) and compound B (ethylene) before they react.
The total pressure in the system is the sum of the partial pressures of A and B. The partial pressure of A can be calculated using the ideal gas law equation:
P(A) = (n(A) × R × T) / V
Substituting the values:
P(A) = (0.48094 mol × 0.0821 L·atm/(mol·K) × 273.15 K) / 10.0 L = 10.62 atm
The partial pressure of B can be calculated in the same way:
P(B) = (n(B) × R × T) / V
Substituting the values:
P(B) = (0.41888 mol × 0.0821 L·atm/(mol·K) × 273.15 K) / 8.60 L = 11.147 atm
Finally, the total pressure in the system is the sum of P(A) and P(B):
Total pressure = P(A) + P(B) = 10.62 atm + 11.147 atm = 21.767 atm.
Therefore, the total pressure in the system will be approximately 21.767 atm.