Gases at nonstandard conditions... HELP!?

Question

Chlorine (Cl2) is produced by the reaction of manganese dioxide (MnO2) and hydrochloric acid (HCl) according to the equation:

MnO2(s) + 4HCl(aq) --> MnCl2(aq) + 2H2O(l) + Cl2(g)

1. How many grams of HCl are required to produce 224 cubic decimeters of Cl2 at 27 degrees Celsius and 100 kPa?

2. How many cubic decimeters of chlorine (Cl2) at 20 degrees Celsius and 102 kPa will be produced from 1000g of manganese dioxide (MnO2)?

Turpentine (C10H16) burns in chlorine (Cl2) to produce carbon (C) and hydrogen chloride (HCl) according to the Equation:

C10H16(l) + 8Cl2(g) --> 10C(s) + 16HCl(g)

3. How many cubic decimeters of HCl at 100 degrees Celsius and 98.6kPA are produced from 150g of C10H16?

4. If 200 dm^3 of Cl2 at 25 degrees Celsius and 106.6kPa reacts with sufficient C10H16, how many grams of C are produced

Answer 1

Given:

MnO2(s) + 4HCl(aq) --> MnCl2(aq) + 2H2O(l) + Cl2(g)

We need to first convert 224 dm^3 to liters, convert temp to kelvin, and convert kPa to Atm--Then use the ideal gas law:

224 dm^3*1L/1dm^3 = 224 L of chlorine gas

Temp (K) = (273.15 + 27) = 300.15 K

100 kPa * 1 Atm/101.325 kPa = 0.987 atm

Now, Ideal Gas Law:

PV=nRT = 0.987 Atm * 224 L = n * 0.08206 L*Atm/K*mol * 300.15 K

n= 8.97 mol of chlorine gas

Now use stoicheometry along with the balanced equation, shown above, to determine moles of HCl, from which we can get grams of HCl.

8.97 mol Cl2 (g) * 4 mol HCL (aq)/1 mol Cl2 (g) = 35.90 mol HCl (aq)

35.90 mol HCl (aq) * 36.453g HCl/1 mol HCl = 1308 g of HCl (aq)

You get the idea.  Should split these into multiple questions, ya know, more incentive...