A chemical equation is balanced when the

Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.3

AveGali [126]

Answer : The percent yield of $CO_2$ is, 68.4 %

Solution : Given,

Mass of $CH_4$ = 0.16 g

Mass of $O_2$ = 0.84 g

Molar mass of $CH_4$ = 16 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $CH_4$ and $O_2$ .

$\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=\frac{0.16g}{16g/mole}=0.01moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{0.84g}{32g/mole}=0.026moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 0.026 moles of $O_2$ react with $\frac{0.026}{2}=0.013$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$

From the reaction, we conclude that

As, 2 mole of $O_2$ react to give 1 mole of $CO_2$

So, 0.026 moles of $O_2$ react to give $\frac{0.026}{2}=0.013$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2$

$\text{ Mass of }CO_2=(0.013moles)\times (44g/mole)=0.572g$

Theoretical yield of $CO_2$ = 0.572 g

Experimental yield of $CO_2$ = 0.391 g

Now we have to calculate the percent yield of $CO_2$

$\% \text{ yield of }CO_2=\frac{\text{ Experimental yield of }CO_2}{\text{ Theretical yield of }CO_2}\times 100$

$\% \text{ yield of }CO_2=\frac{0.391g}{0.572g}\times 100=68.4\%$

Therefore, the percent yield of $CO_2$ is, 68.4 %

6 0

3 years ago

Not all elements readily fit in the metal or non metal groupings. These elements are called metalloids. Examine the table below

crimeas [40]

Brown luster moderately high conductivity

3 0

3 years ago

Which of the following is an example of a chemical change?

bearhunter [10]

Wait what you answered your own question

6 0

3 years ago

Read 2 more answers

A chemist makes of nickel(II) chloride working solution by adding distilled water to of a stock solution of nickel(II) chloride

Lady_Fox [76]

Answer:

0.0900 mol/L

Explanation:

<em>A chemist makes 330. mL of nickel(II) chloride working solution by adding distilled water to 220. mL of a 0.135 mol/L stock solution of nickel(II) chloride in water. Calculate the concentration of the chemist's working solution. Round your answer to significant digits.</em>

Step 1: Given data

Initial concentration (C₁): 0.135 mol/L

Initial volume (V₁): 220. mL

Final concentration (C₂): ?

Final volume (V₂): 330. mL

Step 2: Calculate the concentration of the final solution

We prepare a dilute solution from a concentrated one. We can calculate the concentration of the working solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁/V₂

C₂ = 0.135 mol/L × 220. mL/330. mL = 0.0900 mol/L

3 0

3 years ago

A mixture of NH3 and N2H4 is placed in a sealed container at 300K. The total pressure is 0.50 atm. The container is heated to 12

kiruha [24]

Answer:

The percent of N2H4 in the original mixture is 25 %

Explanation:

Step 1: Data given

Temperature in the sealed container = 300K

The total pressure = 0.50 atm

The container is heated to 1200K

The total pressure at 1200K = 4.5 atm

Step 2: The balanced equation

2NH3 → N2 + 3H2

N2H4 → N2 + 2H2

Step 3: Calculate the initial moles

p*V = n*R*T

⇒ with p = the total pressure = 0.50 atm

⇒ with V = the volume of the gas

⇒ with n = the total moles of the gasses (n1 + n2)

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature in the container = 300

0.50*V= (n1+n2)*R*300

Step 4: after decmposition,

from 2 moles of NH3 we'll get 4 moles (n1 → 2n1)

from 1 moles of N2H4 we'll get 3 moles. (n2 → 3n2)

The total moles for mixture = 2n1 + 3n2

p*V= n*R*T

⇒ with p = the total pressure at 1200 K = 4.5 atm

⇒ V = The volume

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L* atm/ K*mol

⇒ with T = the temperature = 1200 K

4.5*V = (2n1 + 3n2)*R*1200

0.50*V= (n1+n2)*R*300

Step 5: Calculate the percentage of N2H4

After dividing both equations we get:

n2/(n1+n2) = 1/4

n1 = 3 and n2 = 1

Percentage of N2H4 therfore is => 1*100/4 = 25%

and % of NH3 => 75%

The percent of N2H4 in the original mixture is 25 %

4 0

4 years ago