Consider the following balanced equation: 2N2H4( g )+N2O4( g )3N2( g )+4H2O( g )

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.

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Part A

Complete the first row.

Express your answers using one significant figure separated by commas.

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Part B

Complete the second row.

Express your answers as integers separated by commas.

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Part C

Complete the third row.

Express your answers using two significant figures separated by commas.

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Part D

Complete the fourth row.

Express your answers using two significant figures separated by commas.

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Part E

Complete the fifth row.

Express your answers using two significant figures separated by commas.

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

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Explanation / Answer

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

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