3rd to 6th row? Exercise 4.6 Consider the balanced equation Part A Complete the
ID: 716580 • Letter: 3
Question
3rd to 6th row? Exercise 4.6 Consider the balanced equation Part A Complete the table showing the appropriate number of moles of reactants and products. When the number of moles of a reactant is provided, fill in the required amount of the other reactant, as well as the moles of each product formed. When the number of moles of a product is provided, fill in the required amount of each reactant to make that amount of product, as well as the amount of the other product that is made Complete the first row Express your answers as integers separated by commas. Mol N,H, Mol N,O4 Mol Na Mol H,O Mol N O4, Mol N, Mol H2O 2 3.5 4 0 Part B 12.7Explanation / Answer
The balanced chemical equation for the reaction is given as
2 N2H4 (g) + N2O4 (g) --------> 3 N2 (g) + 4 H2O (g)
a) As per the stoichiometric equation,
2 moles N2H4 = 1 mole N2O4 = 3 moles N2 = 4 moles H2O.
This completes the first row.
b) Next, turn the things around a bit to complete the second row.
1 mole N2O4 = 2 moles N2H4 = 3 moles N2 = 4 moles H2O.
Therefore,
9 moles N2O4 = (2 moles N2H4)*(9 moles N2O4)/(1 mole N2O4) = (3 moles N2)* (9 moles N2O4)/(1 mole N2O4) = (4 moles H2O)* (9 moles N2O4)/(1 mole N2O4)
====> 9 moles N2O4 = 18 moles N2H4 = 27 moles N2 = 36 moles H2O.
This completes the second row.
c) Again,
4 moles H2O = 3 moles N2 = 1 mole N2O4 = 2 moles N2H4.
Therefore,
22 moles H2O = (3 moles N2)*(22 moles H2O)/(4 moles H2O) = (1 mole N2O4)*(22 moles H2O)/(4 moles H2O) = (2 moles N2H4)* (22 moles H2O)/(4 moles H2O)
====> 22 moles H2O = 16.5 moles N2 = 5.5 moles N2O4 = 11 moles N2H4.
This completes the third row.
d) Again, use the relation in part (a).
3.5 moles N2H4 = (1 moles N2O4)*(3.5 moles N2H4)/(2 moles N2H4) = (3 moles N2)* (3.5 moles N2H4)/(2 moles N2H4) = (4 moles H2O)* (3.5 moles N2H4)/(2 moles N2H4)
====> 3.5 moles N2H4 = 1.75 moles N2H4 = 5.25 moles N2 = 7.0 moles H2O.
This completes the fourth row.
e) 4.0 moles N2O4 = (2 moles N2H4)*(4.0 moles N2O4)/(1 mole N2O4) = (3 moles N2)* (4.0 moles N2O4)/(1 mole N2O4) = (4 moles H2O)* (4.0 moles N2O4)/(1 mole N2O4)
====> 4.0 moles N2O4 = 8.0 moles N2H4 = 12.0 moles N2 = 16 moles H2O.
This completes the fifth row.
f) Next, use the relation
3 moles N2 = 2 moles N2H4 = 1 mole N2O4 = 4 moles H2O.
Therefore,
12.7 moles N2 = (2 moles N2H4)*(12.7 moles N2)/(3 moles N2) = (1 mole N2O4)* (12.7 moles N2)/(3 moles N2) = (4 moles H2O)* (12.7 moles N2)/(3 moles N2)
====> 12.7 moles N2 = 8.47 moles N2H4 = 4.23 moles N2O4 = 16.93 moles H2O
====> 12.7 moles N2 = 8.5 moles N2H4 = 4.2 moles N2O4 = 16.9 moles H2O
This completes the 6th. row.
Fill in the table as below.
Mol N2H4
Mol N2O4
Mol N2
Mol H2O
2
1
3
4
18
9
27
36
11
5.5
16.5
22
3.5
1.75
5.25
7
8.0
4.0
12.0
16.0
8.5
4.2
12.7
16.9
Mol N2H4
Mol N2O4
Mol N2
Mol H2O
2
1
3
4
18
9
27
36
11
5.5
16.5
22
3.5
1.75
5.25
7
8.0
4.0
12.0
16.0
8.5
4.2
12.7
16.9
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