Chemistry Reference
In-Depth Information
Then proceed as before:
Step 1:
2 NaHCO
3
H
2
SO
4
£
Na
2
SO
4
2 CO
2
2 H
2
O
Step 2:
Present initially
0.171
4
0.0499
5
0.000
0.000
Steps 3 and 4:
Change due to
reaction
0.0999
0.0499
5
0.0499
5
0.0999
Step 5:
Present finally
0.071
5
0.000
0.0499
5
0.0999
Change the number of moles of
Na
2
SO
4
to grams:
a
142 g Na
2
SO
4
1 mol Na
2
SO
4
0.0499
5
mol Na
2
SO
4
b
7.09 g Na
2
SO
4
Practice Problem 10.23
What mass of NaBr will be formed by
addition of 12.3 g of
NaHCO
3
to 25.0 g of HBr, both in aqueous solution?
EXAMPLE 10.24
What mass of sodium chlorate will result from the reaction of 15.5 g of aque-
ous sodium carbonate with 42.3 g of aqueous chloric acid? What mass of excess
reagent will remain unreacted?
Solution
First convert the masses to moles:
a
1 mol Na
2
CO
3
106 g Na
2
CO
3
15.5 g Na
2
CO
3
b
0.146
2
mol Na
2
CO
3
present
a
1 mol HClO
3
84.5 g HClO
3
42.3 g HClO
3
b
0.500
6
mol HClO
3
present
Then proceed as before:
Step 1:
2 HClO
3
Na
2
CO
3
£
2 NaClO
3
CO
2
H
2
O
Step 2:
Present initially
0.000
0.000
0.500
6
0.146
2
Steps 3 and 4:
Change due to
reaction
0.292
4
0.146
2
0.292
4
0.146
2
Step 5:
Present finally
0.208
2
0.000
0.282
4
0.146
2
Converting to mass:
a
106.5 g NaClO
3
1 mol NaClO
3
0.292
4
mol NaClO
3
b
31.1 g NaClO
3
a
84.5 g HClO
3
1 mol HClO
3
0.280
2
mol HClO
3
b
23.7 g HClO
3
