All categories

3 years ago

Give the percent yield when 28.16 g of CO2 are formed from the reaction of 8.000 moles of C8H18 with 16.00 moles of O2.

Chemistry

1 answer:

kipiarov [429]3 years ago

7 0

Answer:

Percent yield of CO₂ is 6.25 %.

Explanation:

Given data:

Percent yield of CO₂ = ?

Actual yield of CO₂ = 28.16 g

Number of moles of C₈H₁₈ = 8.000 mol

Number of moles of O₂ = 16.00 mol

Solution:

Chemical equation:

2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Now we will compare the moles of C₈H₁₈ and O₂ with CO₂.

C₈H₁₈ : CO₂

2 : 16

8.000 : 16/2×8.000 = 64 mol

O₂ : CO₂

25 : 16

16 : 16/25×16= 10.24 mol

Less number of moles of CO₂ are produced from 16 moles of O₂. it will limit the yield of CO₂.

Grams of CO₂ produced:

Mass = number of moles × molar mass

Mass = 10.24 mol × 44 g/mol

Mass = 450.56 g

Percentage yield of CO₂:

Percentage yield = actual yield / theoretical yield × 100

Percentage yield = 28.16 g/ 450.56 g× 100

Percentage yield = 6.25 %

You might be interested in

When 0.100 mol of carbon is burned in a closed vessel with8.00

antoniya [11.8K]

Answer : The mass of carbon monoxide form can be 2.8 grams.

Solution : Given,

Moles of C = 0.100 mole

Mass of $O_2$ = 8.00 g

Molar mass of $O_2$ = 32 g/mole

Molar mass of CO = 28 g/mole

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

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2C+O_2\rightarrow 2CO$

From the balanced reaction we conclude that

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

So, 0.1 moles of $C$ react with $\frac{0.1}{2}=0.05$ moles of $O_2$

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

Now we have to calculate the moles of $CO$

From the reaction, we conclude that

As, 2 mole of $C$ react to give 2 mole of $CO$

So, 0.1 moles of $C$ react to give 0.1 moles of $CO$

Now we have to calculate the mass of $CO$

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

$\text{ Mass of }CO=(0.1moles)\times (28g/mole)=2.8g$

Therefore, the mass of carbon monoxide form can be 2.8 grams.

5 0

4 years ago

A 1 liter solution contains 0.247 M nitrous acid and 0.329 M sodium nitrite. Addition of 0.271 moles of calcium hydroxide will:

Inessa05 [86]

Answer:

a. Raise the pH slightly

Explanation:

We know that

Pka of HNO2/KNO2 =3.39

Moles of HNO2 in the buffer=0.247 mol/L×1L=0.247 moles

Moles of NO2-=0.329mol/L×1L=0.329 moles

If 0.271 moles of Ca(OH)2 is added it will neutralise 0.136 moles of acid ,HNO2,remaining HNO2=0.247-0.136=0.111 moles

Moles of NO2- will increase as 0.0333 moles Ca(NO)2 will be formed =0.0333+0.036=0.0693 moles

pH=pka+log [base]/[acid] {henderson -hasselbach equation}

=3.39+log (0.0693/0.0317)=3.39+0.34=3.73

pH=3.73

4 0

3 years ago

What is the advantage of seeds that can be spread over a wide area?

Tema [17]

Answer:

I'm fairly sure it's D

Explanation:

If they're more spread out than they're roots in theory should have more space. Sorry if I'm wrong

6 0

3 years ago

The chemical formula for ferric sulfate is Fe(SO4)3. Determine the following:

Lyrx [107]

Answer :

(a) The number of sulfur atoms are, $31.61\times 10^{23}$ .

(b) The mass of the mass of $Fe_2(SO_4)_3$ is, 1059.682 grams.

(c) The number of moles of $Fe_2(SO_4)_3$ is, $8.63\times 10^{-3}mole$

(d) The mass of the mass of $Fe_2(SO_4)_3$ is, $19.95\times 10^{-22}g$

Explanation :

(a) As we are given the number of moles of $Fe_2(SO_4)_3$ is, 1.75 mole. Now we have to calculate the number of sulfur atoms.

In the $Fe_2(SO_4)_3$ , there are 2 iron atoms, 3 sulfur atoms, 12 oxygen atoms.

As, 1 mole of $Fe_2(SO_4)_3$ contains $3\times 6.022\times 10^{23}$ number of sulfur atoms.

So, 1.75 mole of $Fe_2(SO_4)_3$ contains $1.75\times 3\times 6.022\times 10^{23}=31.61\times 10^{23}$ number of sulfur atoms.

The number of sulfur atoms are, $31.61\times 10^{23}$

(b) As we are given the number of moles of $Fe_2(SO_4)_3$ is, 2.65 mole. Now we have to calculate the mass of $Fe_2(SO_4)_3$ .

$\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3$

The molar mass of $Fe_2(SO_4)_3$ = 399.88 g/mole

$\text{Mass of }Fe_2(SO_4)_3=2.65mole\times 399.88g/mole=1059.682g$

The mass of the mass of $Fe_2(SO_4)_3$ is, 1059.682 grams.

(c) As we are given the mass of $Fe_2(SO_4)_3$ is, 3.45 grams. Now we have to calculate the moles of $Fe_2(SO_4)_3$ .

$\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3$

The molar mass of $Fe_2(SO_4)_3$ = 399.88 g/mole

$3.45g=\text{Moles of }Fe_2(SO_4)_3\times 399.88g/mole$

$\text{Moles of }Fe_2(SO_4)_3=8.63\times 10^{-3}mole$

The number of moles of $Fe_2(SO_4)_3$ is, $8.63\times 10^{-3}mole$

(d) As we are given the formula unit of $Fe_2(SO_4)_3$ is, 3. Now we have to calculate the mass of $Fe_2(SO_4)_3$ .

As we know that 1 mole of $Fe_2(SO_4)_3$ contains $6.022\times 10^{23}$ formula unit.

Formula used :

$\text{Formula unit of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times 6.022\times 10^{23}$

$3=\text{Moles of }Fe_2(SO_4)_3\times 6.022\times 10^{23}$

$\text{Moles of }Fe_2(SO_4)_3=4.989\times 10^{-24}mole$

Now we have to calculate the mass of $Fe_2(SO_4)_3$ .

$\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3$

The molar mass of $Fe_2(SO_4)_3$ = 399.88 g/mole

$\text{Mass of }Fe_2(SO_4)_3=4.989\times 10^{-24}mole\times 399.88g/mole=19.95\times 10^{-22}g$

The mass of the mass of $Fe_2(SO_4)_3$ is, $19.95\times 10^{-22}g$

6 0

3 years ago

What is the pH if the [H+] concentration is 3 x10^-13

Shkiper50 [21]

Answer:

pH = 12.52

Explanation:

Given that,

The [H+] concentration is $3\times 10^{-13}$ .

We need to find its pH.

We know that, the definition of pH is as follows :

$pH=-log[H^+]$

Put all the values,

$pH=-log[3\times 10^{-13}]\\\\pH=12.52$

So, the pH is 12.52.

6 0

3 years ago