Science, asked by korasaki79, 7 months ago

160g of N2H4 +160g of N2O4 ?what is the limiting reagent​

Answers

Answered by coyote94
1

Answer:

Explanation:

Compd. =

m

N

2

H

4

m

N

2

O

4

m

N

2

m

H

2

O

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

m

m

m

l

l

n

=

m

m

l

4

m

m

m

2

m

m

m

6

m

m

l

l

8

m

m

m

l

l

n

=

m

m

l

2

m

m

m

1

m

m

m

3

m

m

l

l

4

m

m

m

l

l

n

=

m

m

l

4

m

m

m

2

m

m

m

6

m

m

l

l

8

m

m

m

l

l

n

=

m

m

l

4.0

m

m

2.0

m

m

l

6.0

m

l

l

8.0

m

m

m

l

l

n

=

m

m

l

1.4

m

m

0.70

m

l

l

2.1

m

l

l

2.8

m

m

m

l

l

n

=

m

l

l

22.2

m

l

l

11.1

m

l

l

33.3

m

44.4

This question involves practice in using molar ratios.

The balanced equation is

2N

2

H

4

+

N

2

O

4

3N

2

+

4H

2

O

Row 1

Moles of N

2

H

4

=

2

mol N

2

O

4

×

2 mol N

2

H

4

1

mol N

2

O

4

=

4 mol N

2

H

4

Moles of N

2

=

2

mol N

2

O

4

×

3 mol N

2

1

mol N

2

O

4

=

6 mol N

2

Moles of H

2

O

=

2

mol N

2

O

4

×

4 mol H

2

O

1

mol N

2

O

4

=

8 mol H

2

O

Do you notice a pattern?

You are using twice as many moles of

N

2

O

4

as in the balanced equation, so you must use twice as many moles of everything else.

The conversion factor is always the molar ratio:

moles of what you want

moles of what you have

.

Row 2

Moles of N

2

O

4

=

2

mol N

2

H

4

×

1 mol N

2

H

4

2

mol N

2

O

4

=

1 mol N

2

H

4

You are using the same number of moles of

N

2

O

4

as in the balanced equation, so you must use the same amount of everything else as in the balanced equation.

You can do the calculation using numbers as above, and you will get

3 mol N

2

and

4 mol H

2

O

.

Row 3

Moles of N

2

H

4

=

8

mol H

2

O

×

2 mol N

2

H

4

4

mol N

2

O

4

=

4 mol N

2

H

4

You are using twice as many moles of

H

2

O

4

as in the balanced equation, so you must use twice as many moles of everything else.

Thus, you will get

2 mol N

2

O

4

and

6 mol N

2

.

Row 4

Row 4 is the same as Row 2, but you are using more precision in your measurements (one more decimal point).

Thus, you get

4.0 mol N

2

H

4

,

6.0 mol N

2

, and

8.0 mol H

2

O

.

Row 5

Moles of N

2

O

4

=

1.4

mol N

2

H

4

×

1 mol N

2

H

4

2

mol N

2

O

4

=

0.70 mol N

2

H

4

You are using 0.70 times as many moles of

N

2

H

4

as in the balanced equation, so you must use 0.70 times as many moles of everything else.

Thus, you will get

2.1 mol N

2

and

2.8 mol H

2

O

.

Row 6

Moles of N

2

H

4

=

33.3

mol N

2

×

2 mol N

2

H

4

3

mol N

2

=

22.2 mol N

2

H

4

You are using 11.1 times as many moles of

N

2

as in the balanced equation, so you must use 11.1 times as many moles of everything else.

Thus, you will get

11.1 mol N

2

O

4

and

44.4 mol H

2

O

.

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