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3 years ago

Write a sentence that descrices how to determine the number of moles of a compound in known mass of a

Chemistry

1 answer:

sattari [20]3 years ago

6 0

Explanation:

<u>Moles is denoted by given mass divided by the molecular mass , </u>

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

For example ,

For a compound X ,

The given mass i.e. w = 20 g

and the molecular mass ,i.e. , m = 10 g / mol

Then the moles can easily be calculated by using the above formula ,

n = w / m

n = 20 g / 10 g/mol = 2 mol

Hence , answer = 2 mol.

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Ethene is converted to ethane by the reaction flows into a catalytic reactor at 25.0 atm and 250.°C with a flow rate of 1050. L/

Sphinxa [80]

Answer : The percent yield of the reaction is, 76.34 %

Explanation : Given,

Pressure of $C_2H_4$ and $H_2$ = 25.0 atm

Temperature of $C_2H_4$ and $H_2$ = $250^oC=273+250=523K$

Volume of $C_2H_4$ = 1050 L per min

Volume of $H_2$ = 1550 L per min

R = gas constant = 0.0821 L.atm/mole.K

Molar mass of $C_2H_6$ = 30 g/mole

First we have to calculate the moles of $C_2H_4$ and $H_2$ by using ideal gas equation.

For $C_2H_4$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1050L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=611.34moles$

For $H_2$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1550L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=902.46moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$C_2H_4+H_2\rightarrow C_2H_6$

From the balanced reaction we conclude that

As, 1 mole of $C_2H_4$ react with 1 mole of $H_2$

So, 611.34 mole of $C_2H_4$ react with 611.34 mole of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $C_2H_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $C_2H_6$ .

As, 1 mole of $C_2H_4$ react to give 1 mole of $C_2H_6$

As, 611.34 mole of $C_2H_4$ react to give 611.34 mole of $C_2H_6$

Now we have to calculate the mass of $C_2H_6$ .

$\text{Mass of }C_2H_6=\text{Moles of }C_2H_6\times \text{Molar mass of }C_2H_6$

$\text{Mass of }C_2H_6=(611.34mole)\times (30g/mole)=18340.2g$

The theoretical yield of $C_2H_6$ = 18340.2 g

The actual yield of $C_2H_6$ = 14.0 kg = 14000 g (1 kg = 1000 g)

Now we have to calculate the percent yield of $C_2H_6$

$\%\text{ yield of }C_2H_6=\frac{\text{Actual yield of }C_2H_6}{\text{Theoretical yield of }C_2H_6}\times 100=\frac{14000g}{18340.2g}\times 100=76.34\%$

Therefore, the percent yield of the reaction is, 76.34 %

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4 years ago

QUESTION:

vredina [299]

Answer:

Explanation:

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3 years ago

Which of the following lists characteristics of a lake and NOT a reservoir A. Artificial dam, fresh water B. natural dam, filled

tester [92]

<span>Which of the following lists characteristics of a lake and NOT a reservoir A. Artificial dam, fresh water B. natural dam, filled by runoff C. earth dam, salt water D. filled by erosion, cement dam
Answer is B: natural dam,filled by runoff</span>

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3 years ago

Read 2 more answers

How many times higher is the concentration of H+ in the Hubbard Brook sample than in unpolluted rainwater?

Anna [14]

Answer:

1. 7 (a neutral solution)

Answer: 10-7= 0.0000001 moles per liter

2. 5.6 (unpolluted rainwater)

Answer: 10-5.6 = 0.0000025 moles per liter

3. 3.7 (first acid rain sample in North America)

Answer: 10-3.7 = 0.00020 moles per liter

The concentration of H+ in the Hubbard Brook sample is 0.00020/0.0000025, which is 80 times higher than the H+ concentration in unpolluted rainwater.

Explanation:

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3 years ago

Serena mixed two samples together: a gray solid that smells like flowers and a gray solid that smells like bleach. She analyzed

MariettaO [177]

Answer:c

Explanation:

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4 years ago