Use the following information to identify element A and

compound 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 11.98 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) 88n 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 carefully 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 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?

And what is your thinking on this.

Element A mass can be determined: you know the mass, the pressure, temp, and volume (you can get the volume from the Nitrogen). You can get the moles of A also.

Element B. You know volume, temp, pressure.That give you moles. You know the mass, and the empirical formula.

This problem is a sequential problem solving. Attack it in order.

THE ANSWER IS:
A. 42.4 g of C2Cl2H4
B. .538 atm

Anonymous,
Can you show how you arrived at your answer?

For part 1:

Use PV = nRT
plug in your information for N2 first.
(1.0395 atm)(x L) = (.88 g N2 x 1 mol N2/28.02g N2) (.08206) (288.15 K)
solve to get that V = .714 L
then, use PV = nRT again to solve for n of the substance A.
(.9803 atm) (.714 L) = (x n )(.08206)(299.15 K) .

you get that there are .0285 mol A
.0285 mol A (x g A/1 mol A) = 2.018 g
x = 70.81 g A
so A is Cl2, because each Cl is 35.45 and Cl is diatomic.