Example 1
solution
Na + H_{2}O = Na(OH) + ½ H_{2}(g)
1.00 g Na | 1 mol Na -------- 23 g Na | ½ mol H_{2} ---------- 1 mol Na | 8.3145 kPa L mol^{-1} K^{-1} 300.0 K ----------------- 100.0 kPa |
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The last factor is realy R T / P used to convert n moles to volume (V = n R T / P).
= 0.542 L |
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Example 2
solution
Ca + H_{2}O = Ca(OH) + H_{2}(g)
1.00 g Ca | 1 mol Ca -------- 40.1 g Ca | 1 mol H_{2} ---------- 1 mol Ca | 8.3145 kPa L mol^{-1} K^{-1} 300.0 K ----------------- 100.0 kPa |
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= 0.622 L |
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Example 3
solution
2 Na + 2 H_{2}O = 2 Na(OH) + H_{2}(g)
Ca + H_{2}O = Ca(OH) + H_{2}(g)
Let x be the mass of Na, then (2.00-x) is the mass of Ca.
We have the following relationship
x g ----- 23.0 g/mol | 1 mol H2 ---- 2 mol Na | + | (2.0 - x) g Ca ------ 40.1 g Ca/mol1 mol | 1 mol H_{2} ---------- 1 mol Ca | = | 1.164 L H_{2} * 100.0 kPa ------------------ 8.3145 kPa L mol^{-1} K^{-1} 300.0 K |
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Simplify to give
x ---- 46.0 | + | 2 ---- 40.1 | - | x ---- 40.1 | = | 0.0467 | all in mol |
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Multiply all terms by (40.1 * 46.0)
40.1 x + 2 * 46.0 - 46.0 x = 86.1
Simplify
-5.9 x = 86.1 - 92.0 = -5.91
Thus,
Mass of Na = x = 1.0 g
Mass of Ca = 2.0 - x = 1.0 g
Mass Percentage of Na = 100* (1 / 2.0) = 50%
Discussion
Mole of Na = 1/23 = 0.0435 mol
Mole percentage = (^{1}/_{23}) / (^{1}/_{23} + ^{1}/_{40.1}) = 0.635 = 63.5%
Compare this example with gravimetric analyses using the reaction
Ag^{+}(aq) + Cl^{-}(aq) = AgCl (s)
where Cl^{-}(aq) comes from the disolution of two salts such as
NaCl and MgCl_{2}.
Also compare with analyses making use of the reaction
Ba^{2+}(aq) + SO_{4}^{2-}(aq) = BaSO_{4} (s)
where the anion SO_{4}^{2-}(aq) comes from the
disolution of two sulfate salts.
Example 4
solution
This indicates that 1.0 L of CO will react with ½ L O_{2}. Assume you have 1.0 L each, the after the reaction, ½ L of O_{2} will remain. Thus, CO is the limiting reagent.
Example 5
solution
1.0 L CO | 1 mol CO_{2} ------ 22.4 L CO | 1 mol CO_{2} ------ 1 mol CO | 100.1 g CaCO_{3} -------------- 1 mol CaCO_{3} |
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= 4.47 g (Theoretical yield) |
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Actual yield = 2.0 g / 4.47 g = 0.45 = 45% yield
Example 6
Solution
IE Na(g) = e^{-} + Na^{+}(g) | | + | EA | Cl(g) + e^{-} = Cl^{-}(g) |H_{sub} | | | | | | |BE |U | | | | | | Na(s) + ½ Cl_{2}(g) = NaCl(s) |
Reaction | E J |
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Na(s) = Na(g) | 108 |
½Cl_{2}(g) = Cl(g) | ½*240 |
Na(g) = e^{-} + Na^{+} | 496 |
Cl(g) + e^{-} = Cl^{-} | -349 |
NaCl(s) = Na(s) + ½ Cl_{2}(g) | 411 |
Sum | |
NaCl(s) = Na^{+}(g) + Cl^{-}(g) | 786 |
Note that only relevant data have been used.