Solution Stoichiometry Worksheet Page 2
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1
2
3
4-4
-4
Initial
1.55 x 10
mol
1.55 x 10
mol
0
0
-4
-4
-4
-4
Change
-1.55 x 10
mol -1.55 x 10
mol
+1.55 x 10
mol
+1.55 x 10
mol
Final
-4
1.55 x 10
mol HCl
-4
= 6.20 x 10
M HCl
0.250 L
DCI14.4.
Given the reaction
2Na(s) + 2H
O(l) → 2NaOH(aq) + H
(g)
2
2
a)
If a piece of sodium weighing 1.25 grams is added to 450 mL of water, calculate the
grams of H
produced.
2
1 mol Na
⎛
⎞
1.25 g Na
23.0 g Na = 0.0543 mol Na
⎜
⎟
⎝
⎠
1.00 g H
O
1 mol H
O
⎛
⎞
⎛
⎞
2
2
450 mL H
O
⎜
O ⎝ ⎜
⎟
⎟
= 25.0 mol H
O
2
2
1 mL H
18 g H
O
⎝
⎠
⎠
2
2
2Na(s)
+
2H
O(l) → 2NaOH(aq) + H
(g)
2
2
Initial
0.0543 mol
25.0 mol
0
0
Change
Final
Clearly Sodium is the limiting reagent when there are 25.0 mole of H
O present initially.
2
2Na(s)
+
2H
O(l) → 2NaOH(aq) + H
(g)
2
2
Initial
0.0543 mol
25.0 mol
0
0
Change -0.0543 mol
-0.0543 mol +0.0543 mol+0.0272 mol
Final
2 mol H
O
⎛
⎞
2
0.0543 mol Na ⎝ ⎜
⎟
= 0.0543 mol H
O reacting
2
2 mol Na
⎠
2 mol NaOH
⎛
⎞
0.0543 mol Na
⎜
⎟
= 0.0543 mol NaOH forming
2 mol Na
⎝
⎠
1 mol H
⎛
⎞
2
⎜
⎟
0.0543 mol Na
2 mol Na = 0.0272 mol H
forming
2
⎝
⎠
2Na(s)
+
2H
O(l) → 2NaOH(aq) +
H
(g)
2
2
Initial
0.0543 mol
25.0 mol
0
0
Change -0.0543 mol
-0.0543 mol +0.0543 mol +0.0272 mol
Final
0 mol
25.0 mol
+0.0543 mol +0.0272 mol
1 mol H
2.02 g
⎛
⎞
2
⎛
⎞
0.0543 mol Na ⎝ ⎜
⎟
= 0.0548 g H
2 mol Na ⎝ ⎜
⎟
2
1 mol H
⎠
⎠
2
CHEM 1314
2
FALL 2005
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