1. Gas Stoichiometry

2. We have looked at stoichiometry: 1) using masses & molar masses, & 2) concentrations. We can use stoichiometry for gas reactions. As before, we need to consider mole ratios when examining reactions quantitatively. At times you will be able to use 22.4 L/mol at STP and 24.8 L/mol at SATP as shortcuts. grams (x)  moles (x)  moles (y)  grams (y) molar mass of y mole ratio from balanced equation molar mass of x P, V, T (x)  P, V, T (y)  PV = nRT

3. Volume to Volume Problem Calculate the volume of CO 2 which will be produced from 8.40 L of O 2 at STP during the combustion of CH 3 OH.

4. Step 1: Write the balanced equation. 2CH 3 OH(aq) + 3O 2 (g) -  2CO 2 (g) + 4 H 2 0(l) What is given? V=8.40L O 2 P= 101.3kPa T=273K

5. Step 2: Determine moles of reactant PV=nRT n= PV/RT n= (101.3kPa)(8.40L)/(8.314 kPaL/molK)(273K) n=0.375mol

6. 2CH 3 OH(l) + 3O 2 (g) -  2CO 2 (g) + 4 H 2 0(g) Step 3: Use the mole ratio to determine the moles of CO 2 (g) x mol of CO 2 = 2 mol of CO 2 0.375mol O 2 3 mol O 2 x = 0.250mol CO 2 0.375mol O 2 X 2 mol of CO 2 = 0.250mol CO 2 3 mol of O 2

7. 2CH 3 OH(l) + 3O 2 (g) -  2CO 2 (g) + 4 H 2 0(g) Step 4: Use moles to determine volume PV=nRT V= nRT/P V= (0.250mol)(8.314 kPaL/molK)(273K)/(101.3kPa) V=5.60L

8. Mass to Volume Example: Calculate the volume of CO 2 which will be produced from 75.3 g of O 2 during the combustion of CH 3 OH at 30°C and 738 mmHg. Step 1. Write the balanced equation. 2CH 3 OH(l) + 3O 2 (g) -  2CO 2 (g) + 4 H 2 0(g)

9. Step 2: What is given? Mass of O 2 – 75.3g Temperature - 30°C (convert to K) = 30 + 273 = 303K Pressure – 738 mmHg (convert to kPa) x kPa = 101.3 kPa 738mmHg 760 mmHg x = 98.4 kPa

10. Step 3: Find the moles of O 2 using the mass. n = m = 75.3 g M 32.00 g/mol = 2.35 mol of O 2

11. Step 4: Use the mole ratio to determine the # of moles of CO 2 x mol CO 2 = 2 mol of CO 2 2.35 mol O 2 3 mol of O 2 = 1.57 mol of CO 2 2.35 mol O 2 X 2 mol of CO 2 =1.57 mol of CO 2 3 mol of O 2

12. Step 5: Use Ideal Gas law to find the volume of CO 2 PV = nRT (98.3 kPa)(V) = (1.57 mol)(8.314)(303K) V = 40.2 L 40.2 L of carbon dioxide will be produced.

13. Example 3: How many litres of H 2 (at STP) will form when 6.71g of Al react with 12.95g of H 2 SO 4 ? Step 1. Write the balanced equation. 2Al(s) + 3H 2 SO 4 (aq)  Al 2 (SO 4 ) 3 (aq) + 3H 2 (g)

14. Step 2: What is given? Mass of Al– 6.71g Mass H 2 SO 4 –12.95g Because given 2 reactants, must be a limiting reactant problem!

15. Step 3: Find the moles of both reactants using the mass. n = m = 6.71 g M 26.98 g/mol =.248 mol of Al n = m = 12.95 g M 98.02 g/mol = 0.132 mol of H 2 SO 4

16. Step 4: Determine limiting reactant using mole ratios.248 mol Al X 3 mol of H 2 SO 4 =.372 mol H 2 SO 4 2 mol of Al Since I need 0.372mol of H 2 SO 4 and I only have 0.132 mole it is the limiting reactant.

17. Step 5: Use the mole ratio to determine the # of moles of H 2 x mol H 2 = 3 mol of H 2 SO 4 0.132 mol H 2 SO 4 3 mol of H 2 = 0.132 mol of H 2 0.132 mol H 2 SO 4 X 3 mol H 2 =0.132 H 2 3 mol H 2 SO 4

18. Step 5: Use Ideal Gas law to find the volume of CO 2 PV = nRT (101.3kPa)(V) = (0.132 mol)(8.314)(273K) V = 2.95 L 2.95 L of hydrogen gas will be produced.

19. Sample problem 4 Hydrogen gas (and NaOH) is produced when sodium metal is added to water. What mass of Na is needed to produce 20.0 L of H 2 at STP? First we need a balanced equation: 2Na(s) + 2H 2 O(l)  H 2 (g) + 2NaOH(aq) 0.893 mol H 2 = 41.1 g Na 2 mol Na 1 mol H 2 x PV = nRT (8.31 kPaL/Kmol )(273 K) (101.3 kPa)(20.0 L) = n = 0.893 mol H 2 P= 101.3 kPa, V= 20.0 L, T= 273 K =1.79mol Na m=23 g/mol x 1.79mol