I posted this before and I know now to use P1T2=P2T1 but I was wondering what I would plug in as pressure and temp? just an example of one would be so great.

I posted this before and I know now to use P1T2=P2T1 but I was wondering what I would plug in as pressure and temp? just an example of one would be so great.

A gas at 760 mm Hg pressure and 25 C is heated to 50 C. What is the new pressure?
p1 = 760 mm Hg
T1 = 25 + 273 = 298 K
P2 = ?
T2 = 50 + 273 = 323 K
Solve for P2.

In each of the four problem in this section, all gases are measured at the same temperature and pressure.

1) (a) What volume of hydrogen will combine with 12L of chlorine to form hydrogen chloride? (b) What volume of hydrogen chloride will be formed?
H2(gas) + Cl2(gas) -> 2HCl(gas)

With all gases we can use a shortcut in which we use liters of gas directly as if they were moles.
H2 + Cl2 ==> 2HCl
12 L Cl2 x (1 mol H2/1 mol Cl2) = 12 L H2 required.

12 L Cl2 x (2 mols HCl/1 mol Cl2) = 12 L x 2/1 = 24 L HCl formed.

2) (a) What volume of hydrogen will react with 6ft^3 of nitrogen to form ammonia? (b) What volume of ammonia will be produced?
N2(gas) + 3H2(gas) -> 2NH3(gas)

Convert 6 ft^3 to L. That's approximately 170 L but you need to get a more accurate answer instead of using this estimate.
170 L N2 x (1 mol N2/3 mols H2) = 170 x 1/ = ? L H2 needed.

170 L N2 x (2 mol NH3/1 mol N2) = 170 x 2/1 = ?L NH3 formed.

3) Sixty-four liters of NO are mixed with 40L of O2 and allowed to react. What is the total volume of gas present after completion of the reaction?
2NO(gas) + O2(gas) -> 2NO2(gas)

First determine the limiting reagent.If 64 is the limiting reagent it will be
64 L NO x (2 mols NO2/2 mols NO) = 64 L NO2 formed.
If 40 L O2 is the limiting reagent it will be
40 L O2 x (2 mols NO2/1 molO2) = 80 L NO2 formed.
In limiting reagent (LR) problems the small number is always the winner; therefore, N2 is the LR and O2 is the excess reagent. We start there.
So how much O2 is used to use up all of the N2. That's
64 L N2 x (1 mol O2/2 mol N2) = 64 x 1/2 = 32 L O2 used. We had 40 so that means 8 L O2 is left unreacted. Now we have zero L N2, 8 L O2 and 64 L NO2 for a total volume of ? L.

4) (a) What volume of O2 at S.T.P if required for the complete combustion of 1 mol of carbon disulfide, CS2? (b) What volumes of CO2 and SO2 are produced (also at S.T.P.)?
CS2(liquid) + 3O2(gas) -> CO2(gas) + 2SO2(gas)

1 mol CS2 x (3 mols O2/1 mol CS2) - 1 mol CS2 x 3/1 = ? mols CS2 needed. Then remember that a gas occupies 22.4 L at STP.
To find moles and volume of CO2 and SO2 just follow regular stoichiometry as I've done above. Post your work if you have trouble.