How many bonding electrons are between the two oxygen atoms in hydrogen peroxide, h2o2h2o2?

How many moles of carbon dioxide will be produced from the complete combustion of 13.6 moles of butane?

Aleks04 [339]

Answer:

54.4 mol

Explanation:

the equation for complete combustion of butane is

2C₄H₁₀ + 13O₂ ---> 8CO₂ + 10H₂O

molar ratio of butane to CO₂ is 2:8

this means that for every 2 mol of butane that reacts with excess oxygen, 8 mol of CO₂ is produced

when 2 mol of C₄H₁₀ reacts - 8 mol of CO₂ is produced

therefore when 13.6 mol of C₄H₁₀ reacts - 8/2 x 13.6 mol = 54.4 mol of CO₂ is produced

therefore 54.4 mol of CO₂ is produced

5 0

3 years ago

How many sulfur atoms are expected in the reactants of this chemical reaction?

Rudik [331]

Answer is B, because CaSO3 only has one S

4 0

3 years ago

Which one is the catalyst?

yan [13]

Answer:

the catalyst is the two gray dots

5 0

3 years ago

Check all that apply to sugar.

harina [27]

Answer:

Candy

Explanation:

its sweet and has a lot of sugar and acid

3 0

3 years ago

Read 2 more answers

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