In a reaction, 2mol of oxygen reacts with 8g of ethane to produce carbon dioxide and water, calculate the mass of carbon dioxide formed in this reaction. 2C2H6 +7O2 > 4CO2 +6H2O

Answers

Answer 1

Answer:

The mass of carbon dioxide formed in the reaction is 23.47 grams.

Explanation:

The balanced reaction is:

2 C₂H₆ +7 O₂ ⇒ 4 CO₂ + 6 H₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

C₂H₆: 2 moles O₂: 7 moles CO₂: 4 molesH₂O: 6 moles

Being the molar mass of the compounds:

C₂H₆: 30 g/moleO₂: 16 g/mole CO₂: 44 g/moleH₂O: 18 g/mole

By reaction stoichiometry, the following mass amounts of each compound participate in the reaction:

C₂H₆: 2 moles* 30 g/mole= 60 gO₂: 7 moles* 16 g/mole= 112 gCO₂: 4 moles* 44 g/mole= 176 gH₂O: 6 moles* 18 g/mole= 108 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows:  if by stoichiometry of the reaction 7 moles of O₂ react with 60 g of C₂H₆, 2 moles of O₂ will react with how much mass of C₂H₆?

[tex]mass of C_{2}H_{6} =\frac{2 moles of O_{2}* 60 grams of C_{2}H_{6} }{7 moles of O_{2}}[/tex]

mass of C₂H₆= 17.14 grams

But 17.14 grams of C₂H₆ are not available, 8 grams are available. Since you have less moles than you need to react with 2 moles of O₂, ethane C₂H₆ will be the limiting reagent.

Then, it is possible to determine the amount of mass of CO₂ produced by another rule of three: if by stoichiometry 60 grams of C₂H₆ produce 176 grams of CO₂, how many mass of CO₂ will be formed if 8 grams of C₂H₆ react?

[tex]mass of CO_{2} =\frac{8 grams of C_{2}H_{6}* 176 grams of CO_{2} }{60 grams of C_{2}H_{6}}[/tex]

mass of CO₂= 23.47 grams

The mass of carbon dioxide formed in the reaction is 23.47 grams.


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How many moles of butane (C4H10(g)) were combusted in the presence of excess
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Answers

Moles of Butane combusted : 0.4982

Further explanation

Given

55.8 water vapour at STP

Required

moles of Butane

Solution

Reaction(Combustion of Butane) :

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

1 mol at STP = 22.4 L, so mol of H₂O for 55.8 L :

=55.8 : 22.4

=2.491 moles

mol H₂O based on Butane as a limiting reactant(excess  Oxygen)

From the equation, mol ratio Butane-C₄H₁₀ : H₂O = 2 : 10, so mol Butane :

[tex]\tt =\dfrac{2}{10}\times 2.491\\\\=0.4982[/tex]

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Answers

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The chemical energy in a substance is determined by the potential energy of the bonds of the products formed when it reacts

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mg(oh)2
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Answers

Answer:

A

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Which of the following is not true about one mole?

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1 mole of boron atoms
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the number of atoms in 1 mole of carbon = the number of atoms in 1 mole of boron

12 g of carbon equals one mole of carbon atoms

Answers

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A

Explanation:

Bc ik

The number of atoms in 1 mole of carbon = the number of atoms in 1 mole of boron is not true about one mole.

What element of one mole is false?

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Given that the molar masses of carbon and boron differ, the statement that the number of atoms in a mole of carbon is equal to the number of atoms in a mole of boron is false.

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The complete question is:

Which of the following is not true about one mole?

a) the mass of 1 mole of carbon atoms = the mass of

b) 1 mole of boron atoms

one mole contains 6.02 x 1023 particles

c) the number of atoms in 1 mole of carbon = the number of atoms in 1 mole of boron

d) 12 g of carbon equals one mole of carbon atoms

To learn more about one mole refer to:

https://brainly.com/question/28037816

#SPJ2

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Answers

Answer : The molecular formula of benzoyl peroxide is [tex]C_{14}H_{10}O_4[/tex]

Explanation :

Empirical formula : It is the simplest form of the chemical formula which depicts the whole number of atoms of each element present in the compound.  

Molecular formula : it is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

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