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Solving Limiting Reactant Problems

Solving Limiting Reactant Problems. Background. In limiting reactant problems , we have the amounts (masses or mols) of two of the reactants. The problem is always to find out which one of the reactants is “ limiting .” The limiting reactant is the one that we will run out of first.

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Solving Limiting Reactant Problems

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  1. Solving Limiting Reactant Problems

  2. Background • In limiting reactant problems, we have the amounts (masses or mols) of two of the reactants. • The problem is always to find out which one of the reactants is “limiting.” • The limiting reactant is the one that we will run out of first. • The other reactant will be in excess and will not be totally used up in the reaction.

  3. Background • We will be using the same techniques we use to solve ideal stoichiometric problems. • We will use molar masses and mole ratios to determine the limiting reactant. • We will also use these values to figure out how much product we can expect from that limiting reactant.

  4. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l)

  5. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) First, we list the molar masses of the compounds.

  6. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) First, we list the molar masses of the compounds.

  7. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) Next, put in our known values and our unknown quantity.

  8. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) m (g) 155 350 ( ? ) Next, put in our known values and our unknown quantity.

  9. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) m (g) 155 350 ( ? ) Next, calculate the number of moles of the knowns.

  10. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) m (g) 155 350 ( ? ) n (mol) Next, calculate the number of moles of the knowns.

  11. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) m (g) 155 350 ( ? ) n (mol) mC5H12 155 g nC5H12 = = = 2.15 mol MC5H12 72.0 g/mol

  12. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) m (g) 155 350 ( ? ) n (mol) 2.15 mC5H12 155 g nC5H12 = = = 2.15 mol MC5H12 72.0 g/mol

  13. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) m (g) 155 350 ( ? ) n (mol) 2.15 mO2 350 g nO2 = = = 10.94 mol MO2 32.0 g/mol

  14. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) m (g) 155 350 ( ? ) n (mol) 2.15 10.94 mO2 350 g nO2 = = = 10.94 mol MO2 32.0 g/mol

  15. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) m (g) 155 350 ( ? ) n (mol) 2.15 10.94 If we used up all 2.15 mols of the C5H12, we would need 8×2.15 mol = 17.20 mols of O2.

  16. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) m (g) 155 350 ( ? ) n (mol) 2.15 10.94 We only have 10.94 mols of O2. Therefore, O2 is the limiting reactant.

  17. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) m (g) 155 350 ( ? ) n (mol) 2.15 10.94 Now, we can calculate the number of mols of CO2 produced.

  18. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) m (g) 155 350 ( ? ) n (mol) 2.15 10.94 coefficient CO2 5 nCO2 = nO2 × = 10.94 mol × = 6.84 mol coefficient O28

  19. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) m (g) 155 350 ( ? ) n (mol) 2.15 10.94 6.84 coefficient CO2 5 nCO2 = nO2 × = 10.94 mol × = 6.84 mol coefficient O28

  20. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) m (g) 155 350 ( ? ) n (mol) 2.15 10.94 6.84 Finally, we can calculate the amount of CO2 produced.

  21. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) m (g) 155 350 ( ? ) n (mol) 2.15 10.94 6.84 mCO2 = nCO2 × MCO2 = (6.84 mol)(44.0 g/mol) = 301 g

  22. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) m (g) 155 350 301 n (mol) 2.15 10.94 6.84 mCO2 = nCO2 × MCO2 = (6.84 mol)(44.0 g/mol) = 301 g

  23. Example • How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C5H12, with 350 g of oxygen gas, O2? M (g/mol) 72.0 32.0 44.0 18.0 C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l) m (g) 155 350 301 n (mol) 2.15 10.94 6.84 mCO2 = nCO2 × MCO2 = (6.84 mol)(44.0 g/mol) = 301 g

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