Ba(ClO 4 ) 2 (aq)+K 2 SO 4 (aq) -> BaSO 4 (s)+2 KClO 4 (aq)

How many grams of CO2 are used when 7 g of oxygen are produced

lara [203]

308 grams of co2 is used

4 0

3 years ago

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T

Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 730.5 mmHg

V = Volume of the gas = 9.49 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol$

According to the reaction shown below as:-

$2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when $\frac{2}{3}\times 0.3794$ moles of potassium chlorate undergoes reaction.

<u>Moles of potassium chlorate reacted = 0.2529 moles</u>

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 735.5 mmHg

V = Volume of the gas = 9.99 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol$

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g$

<u>Hence, the amount of oxygen gas collected will be 12.8675 g</u>

5 0

3 years ago

4 Nitrogen monoxide reacts with oxygen like this:

olga2289 [7]

a. mol O₂=0.5

b. volume O₂ = 25 cm³

c. i. the total volume of the two reactants = 75 cm³

c. ii. the volume of nitrogen dioxide formed = 50 cm³

<h3>Further explanation</h3>

Reaction

2NO(gas) + O₂(gas) ⇒ 2NO₂ (gas)

a.

mol NO = 1

From the equation, mol ratio NO : O₂ = 2 : 1, so mol O₂ :

$\tt \dfrac{1}{2}\times 1=0.5$

b.

From Avogadro's hypothesis, at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles

Because mol ratio NO : O₂ = 2 : 1, so volume O₂ :

$\tt \dfrac{1}{2}\times 50~cm^3=25~cm^3$

c.

i. total volume of reactants : 25 cm³+ 50 cm³=75 cm³

ii. the volume of nitrogen dioxide formed :

mol ratio NO : NO₂ = 2 : 2, so volume NO₂ = volume NO = 50 cm³

6 0

3 years ago

An aluminum cylinder has a radius of 3.2 cm and a length of 8.6 cm. What is the mass of the aluminum cylinder if aluminum has a

Hatshy [7]

Answer:

750g of Aluminum

Assuming 8.6 cm lenth is the height of the cylinder. The volume of a cylinder is: V = $\pi$ *r^2*h

V = 3.14cm x 10.24cm x 3.6cm

V = 280cm^3

Now density = mass/volume

2.7g/cm^3 = mass/280cm^3

2.7g/cm^3 x 280cm^3 = mass/<u>280cm^3</u> x <u>280cm^3</u>

= 750g of Aluminum

5 0

3 years ago

Compound X has the formula C8H14. X reacts with one molar equivalent of hydrogen in the presence of a palladium catalyst to form

REY [17]

Answer:

1-Ethyl-3-methylidenecyclopentane

Step-by-step explanation:

Formula = C₈H₁₄. An alkane has formula C₈H₁₈. ∴ X contains 2 double bonds, 2 rings, or 1 ring and 1 double bond.

X absorbs only 1 mol of hydrogen. ∴ X contains 1 ring and 1 double bond.

Hydrogenation gives 1-ethyl-3-methylcyclopentane.

Ozonolysis gives formaldehyde, so X must contain a =CH₂ group.

Hydrogenation of X converted the =CH₂ to -CH₃.

X is 1-ethyl-3-methylidenecyclopentane.

You can see the reactions in the image below.

3 0

3 years ago