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MrMuchimi
3 years ago
11

A tank at is filled with of sulfur tetrafluoride gas and of sulfur hexafluoride gas. You can assume both gases behave as ideal g

ases under these conditions. Calculate the mole fraction of each gas. Round each of your answers to significant digits.
Chemistry
1 answer:
dangina [55]3 years ago
7 0

The question is incomplete, the complete question is:

A 7.00 L tank at 21.4^oC is filled with 5.43 g of sulfur hexafluoride gas and 14.2 g of sulfur tetrafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas. Round each of your answers to significant digits.

<u>Answer:</u> The mole fraction of sulfur hexafluoride is 0.221 and that of sulfur tetrafluoride is 0.779

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

  • <u>For sulfur hexafluoride:</u>

Given mass of sulfur hexafluoride = 5.43 g

Molar mass of sulfur hexafluoride = 146.06 g/mol

Putting values in equation 1, we get:

\text{Moles of sulfur hexafluoride}=\frac{5.43g}{146.06g/mol}=0.0372mol

  • <u>For sulfur tetrafluoride:</u>

Given mass of sulfur tetrafluoride = 14.2 g

Molar mass of sulfur tetrafluoride = 108.07 g/mol

Putting values in equation 1, we get:

\text{Moles of sulfur tetrafluoride }=\frac{14.2g}{108.07g/mol}=0.1314mol

Total moles of gas in the tank = [0.0372+ 0.1314] mol = 0.1686 mol

Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:

\chi_A=\frac{n_A}{n_A+n_B} .....(2)

where n is the number of moles

Putting values in equation 2, we get:

\chi_{SF_6}=\frac{0.0372}{0.1686}=0.221

\chi_{SF_4}=\frac{0.1314}{0.1686}=0.779

Hence, the mole fraction of sulfur hexafluoride is 0.221 and that of sulfur tetrafluoride is 0.779

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insens350 [35]

Answer:

Explanation:

The first thing that you need to do here is to calculate the theoretical yield of the reaction, i.e. what you get if the reaction has a

100

%

yield.

The balanced chemical equation

N

2

(

g

)

+

3

H

2

(

g

)

→

2

NH

3

(

g

)

tells you that every

1

mole of nitrogen gas that takes part in the reaction will consume

3

moles of hydrogen gas and produce

1

mole of ammonia.

In your case, you know that

1

mole of nitrogen gas reacts with

1

mole of hydrogen gas. Since you don't have enough hydrogen gas to ensure that all the moles of nitrogen gas can react

what you need

3 moles H (sub 2)

>

what you have

1 mole H (sub2)

you can say that hydrogen gas will act as a limiting reagent, i.e. it will be completely consumed before all the moles of nitrogen gas will get the chance to take part in the reaction.

So, the reaction will consume

1

mole of hydrogen gas and produce

1

mole H

2

⋅

2 moles NH

3

3

moles H

2

=

0.667 moles NH

3

at

100

%

yield. This represents the reaction's theoretical yield.

Now, you know that the reaction produced

0.50

moles of ammonia. This represents the reaction's actual yield.

In order to find the percent yield, you need to figure out how many moles of ammonia are actually produced for every

100

moles of ammonia that could theoretically be produced.

You know that

0.667

moles will produce

0.50

moles, so you can say that

100

moles NH

3

.

in theory

⋅

0.50 moles NH

3

.

actual

0.667

moles NH

3

.

in theory

=

75 moles NH

3

.

actual

Therefore, you can say that the reaction has a percent yield equal to

% yield = 75%

−−−−−−−−−−−−−

or 75 moles NH sub3

I'll leave the answer rounded to two sig figs.

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