All categories

3 years ago

Consider the reaction 2 CO + O2 → 2 CO2 .What is the percent yield of carbon dioxide

Chemistry

1 answer:

Wewaii [24]3 years ago

7 0

Answer:

$Y=50.9\%$

Explanation:

Hello,

In this case, given the reaction, we can directly compute the theoretically yielded grams of carbon dioxide, considering the 2:2 molar ratio between carbon monoxide (molar mass = 28 g/mol) and carbon dioxide (molar mass = 44 g/mol) and the initial reacting grams of carbon monoxide in excess oxygen:

$m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO}*\frac{44gCO_2}{1molCO_2} =15.7gCO_2$

Thus, as only 8 g were actually yielded, we compute the percent yield:

$Y=\frac{8g}{15.7g}*100\% \\\\Y=50.9\%$

Best regards.

You might be interested in

How is science used on a daily basis give 3 reasons how

goldenfox [79]

1-Autonomy: Our desire to direct our own lives. In short: “You probably want to do something interesting, let me get out of your way!”
2-Mastery: Our urge to get better at stuff.
3-Purpose: The feeling and intention that we can make a difference in the world.

7 0

4 years ago

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$

From the reaction, we conclude that

As, 2 mole of $Na$ react to give 2 mole of $NaCl$

So, 0.217 mole of $HCl$ react to give 0.217 mole of $NaCl$

Now we have to calculate the mass of $NaCl$

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

Molar mass of $NaCl$ = 58.5 g/mole

$\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g$

Now we have to calculate the percent yield of this reaction.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = $\frac{5.24g}{12.7g}\times 100$

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

4 0

4 years ago

Can the crabs see the plankton they eat near the ocean floor?

cricket20 [7]

Answer:

Explanation:

The crabs cannot see the plankton they eat near the ocean floor. For the crabs to see the plankton, some color of visible light would need to reach the plankton so that it can be reflected into the crabs' eyes.

4 0

3 years ago

Read 2 more answers

You need to make 10.0 L of 1.2 M KNO3. What molar ( concentration) would the potassium nitrate solution need to be if you were t

solniwko [45]

Answer:

2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.

Explanation:

Generally, moles of solute in solution before dilution must equal moles of solute after dilution.

By definition Molarity = moles solute/volume of solution in Liters

=> moles solute = Molarity x Volume (L)

Apply moles before dilution = moles after dilution ...

=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution

=> (M)(2.5L)before = (1.2M)(10.0L)after

=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate

6 0

3 years ago

Calculate the number of moles of ethanol acid in 30 cm³of 1 mol dm-³ solution

muminat

Answer:

May I assume "ethanol acid is just ethanol (it has one slightly acidic H atom). If so, the molar mass is 46.02 g/mole.

Explanation:

We have 30 cm^3 [30 ml] of 1.0 M (1 mole/liter) [1 dm³ = 1 liter].

That is 1 mole/liter. 30 ml would contain (0.030 liter)*(1 mole/1 liter) = 0.03 moles.

7 0

3 years ago