Solution to Exercises for Study Sheet 4
1. Each balanced equation is provided in Appendix X.
2.
2 Al (s)
MM
27.0
+
6 HClO4
(aq)
→
100.5
2 Al(ClO4)3 (aq)
325.3
+
3 H2 (g)
2.02
⎛ 6 mol HClO4 ⎞
a) ? mol HClO4 = 0.400 mol H 2 ⎜
= 0.800 mol HClO4
⎝ 3 mol H 2 ⎟⎠
⎛ 1 mol Al ⎞
b) ? g HClO4 = 50.0 g Al ⎜
⎝ 27.0 g Al ⎟⎠
⎛ 6 mol HClO4 ⎞ ⎛ 100. g HClO4 ⎞
⎜⎝ 2 mol Al ⎟⎠ ⎜⎝ 1 mol HClO ⎟⎠ = 558. g HClO4
4
⎛ 2 mol Al(ClO4 )3 ⎞ ⎛ 325.3 g Al(ClO4 )3 ⎞
c) ? g Al(ClO4 )3 = 1.40 mol H 2 ⎜
⎟⎠ ⎜⎝ 1 mol Al(ClO ) ⎟⎠ = 304. g Al(ClO4 )3
3 mol H 2
⎝
4 3
⎛ 1 mol Al ⎞ ⎛ 3 mol H 2 ⎞ ⎛ 2.02 g H 2 ⎞
d) ? g H 2 = 25.0 g Al ⎜
⎟⎜
⎟⎜
⎟ = 2.81 g H 2
Excess ⎝ 27.0 g Al ⎠ ⎝ 2 mol Al ⎠ ⎝ 1 mol H 2 ⎠
Reagent
⎛ 1 mol HClO4 ⎞ ⎛ 3 mol H 2 ⎞ ⎛ 2.02 g H 2 ⎞
? g H 2 = 150 g HClO4 ⎜
= 1.51 g H 2
⎝ 100.5 g HClO4 ⎟⎠ ⎜⎝ 6 mol HClO4 ⎟⎠ ⎜⎝ 1 mol H 2 ⎟⎠
Limiting
Reagent
⎛ 1 mol HClO4 ⎞ ⎛ 2 mol Al ⎞ ⎛ 27.0 g Al ⎞
e) ? g Al = 150 g HClO4 ⎜
⎟⎠ ⎜⎝ 6 mol HClO ⎟⎠ ⎜⎝ 1 mol Al ⎟⎠ = 13.4 g Al
100.5
g
HClO
⎝
4
4
(only needed)
Start with
Limiting
Reagent
? g Al (in excess) = g Al (initial) − g Al (only needed) = 25.0 g − 13.4 g = 11.6 g
Limiting
reagent
problem