Question

Consider the following balanced equation: 2N2H4(g)+N2O4(g)→3N2(g)+4H2O(g)Complete the following table showing the appropriate number of moles of reactants and products. If 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. If 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.

Mol N2H4	Mol N2O4	Mol N2	Mol H2O
4	_____	_____	_____
_____	7	_____	_____
_____	_____	_____	18
3.5	_____	_____	_____
_____	3.6	_____	____
_____	_____	12.7	_____

Part A

Complete the first row.

Express your answers using one significant figure separated by commas.

Mol N2O4, Mol N2, Mol H2O =

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Part B

Complete the second row.

Express your answers as integers separated by commas.

Mol N2H4, Mol N2, Mol H2O =

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Part C

Complete the third row.

Express your answers using two significant figures separated by commas.

Mol N2H4, Mol N2O4, Mol N2 =

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Part D

Complete the fourth row.

Express your answers using two significant figures separated by commas.

Mol N2O4, Mol N2, Mol H2O =

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Part E

Complete the fifth row.

Express your answers using two significant figures separated by commas.

Mol N2H4, Mol N2, Mol H2O =

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Part F

Complete the sixth row.

Express your answers using three significant figures separated by commas.

Mol N2H4, Mol N2O4, Mol H2O=

Answer 1

the given reaction is

2N2H4 + N204 ----> 3N2 + 4H20

1)

given 4 moles of N2H4

we can see that

moles of N204 = 0.5 x moles of N2H4 = 0.5 x 4 = 2

moles of N2 = 3 x moles of N204 = 2 x 3 = 6

moles of H20 = 2 x moles of N2H4 = 2 x 4 = 8

2)

given 7 moles of N204

we can see that

moles of N2H4 = 2 x moles of N204 = 2 x 7 = 14

moles of N2 = 3 x moles of N204 = 3 x 7 = 21

moles of H20 = 4 x moles of N204 = 4 x 7 = 28

3)

given 18 mol of H20

we can see that

moles of N2H4 = 0.5 x moles of H20 = 0.5 x 18 = 9

moles of N204 = 0.25 x moles of H20 = 0.25 x 18 = 4.5

moles of N2 = 3 x moles of N204 = 3 x 4.5 = 13.5

4)

given 3.5 moles of N2H4

we can see that

moles of N204 = 0.5 x moles of N2H4 = 0.5 x 3.5 = 1.75

moles of N2 = 3 x moles of N204 = 1.73 x 3 = 5.25

moles of H20 = 2 x moles of N2H4 = 2 x 3.5 = 7

5)

given 3.6 moles of N204

we can see that

moles of N2H4 = 2 x moles of N204 = 2 x 3.6 = 7.2

moles of N2 = 3 x moles of N204 = 3 x 3.6 = 10.8

moles of H20 = 4 x moles of N204 = 4 x 3.6 = 14.4

6) given 12.7 mol of N2

we can see that

moles of N204 = (1/3) x moles of N2 = (1/3) x 12.7 = 4.2333

moles of N2H4 = (2/3) x moles of N2 = 2 x 12.7 / 3 = 8.4666

moles of H20 = (4/3) x moles of N2 = 4 x 12.7 / 3 = 16.933