C2H6O(g) + 3O2(g) 2CO2(g) + 3H2O(g) Inside the piston of an automobile engine, 0.461 g of ethanol (C2H6O) gas reacts with 0.640 grams of oxygen gas to produce carbon dioxide and water, according to the balanced equation shown above. What is the limiting reagent?

Answers

Answer 1

Answer:

Oxygen is the limiting reactant.

Explanation:

Hello,

In this case, given the reaction:

[tex]C_2H_6O(g) + 3O_2(g)\rightarrow 2CO_2(g) + 3H_2O(g)[/tex]

Hence, given the masses of both ethanol and oxygen, we are able to compute the available moles ethanol by:

[tex]n_{C_2H_6O}^{available}=0.461g*\frac{1mol}{46g}=0.01mol C_2H_6O[/tex]

Next, we compute the moles of ethanol that react with the 0.640 grams of oxygen considering their 1:3 molar ratio in the chemical reaction:

[tex]n_{C_2H_6O}^{consumed\ by\ O_2}=0.64gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6O}{3molO_2} =0.0067molC_2H_6O[/tex]

In such a way, since there are 0.01 available moles of ethanol but just 0.0067 moles are reacting, we evidence ethanol is in excess, therefore the oxygen is the limiting reactant.

Best regards.

Answer 2

Answer:

Oxygen O₂ will be the limiting reagent.

Explanation:

C₂H₆O(g) + 3 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(g)

First of all you must know by stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction) how much mass of each compound reacts. First of all, being:

C: 12 g/moleH: 1 g/moleO: 16 g/mole

the molar mass of each reagent is:

C₂H₆O: 2*12 g/mole + 6* 1 g/mole + 16 g/mole= 46 g/moleO₂: 2*16 g/mole= 32 g/mole

Then, since 1 mol of C₂H₆O and 3 moles of O₂ react by stoichiometry, the amount of mass that reacts is:

C₂H₆O: 1 mole* 46 g/mole= 46 gO₂: 3* 32 g/mole= 96 g

Now you apply a rule of three as follows: if 46 g of C₂H₆O reacts with 96 g of O₂ by stoichiometry, 0.461 g of C₂H₆O with how much mass of O₂ will they react?

[tex]mass of O_{2} =\frac{0.461 grams of C_{2}H_{6} O*96 grams ofO_{2} }{46 grams of C_{2}H_{6} O}[/tex]

mass of O₂= 0.962 grams

But 0.962 grams of O₂ are not available, 0.640 grams are available. Since you have less mass than you need to react with 0.461 g of C₂H₆O, oxygen O₂ will be the limiting reagent.


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Answers

Answer:

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The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.
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Calculate the volume of dry CO2 produced at body temperature (37°C) and 0.960 atm when 24.5 g of glucose is consumed in this reaction.

Answers

Answer:

[tex]\large \boxed{\text{21.6 L}}[/tex]

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

         C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g:      24.5

(a) Moles of C₆H₁₂O₆

[tex]\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}[/tex]

(b) Moles of CO₂

[tex]\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}[/tex]

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

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p = 0.960 atm

n = 0.8159 mol

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(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

[tex]\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}[/tex]

You are given the reaction Cu + HNO3 Cu(NO3)2 + NO + H2O complete the final balanced equation based on half-reactions

Answers

The balanced equation based on half-reactions is 2Cu + 8HNO3 + 6H+ -> 2Cu(NO3)2 + 2NO + 4H2O.

How to determine the balanced equation based on half-reactions

To complete the balanced equation for the given reaction Cu + HNO3 -> Cu(NO3)2 + NO + H2O using half-reactions, we need to break down the overall reaction into separate oxidation and reduction half-reactions.

1. Oxidation Half-Reaction:

Cu -> Cu2+ + 2e-

In this step, copper (Cu) is oxidized, losing two electrons to form copper(II) ions (Cu2+).

2. Reduction Half-Reaction:

HNO3 + 3H+ + 2e- -> NO + 2H2O

In this step, nitric acid (HNO3) is reduced, gaining two electrons to form nitric oxide (NO) and water (H2O).

Now, to balance the half-reactions, we need to make sure that the number of electrons gained in the reduction half-reaction is equal to the number of electrons lost in the oxidation half-reaction. In this case, we can achieve this by multiplying the oxidation half-reaction by two.

Balanced Half-Reactions:

Oxidation: 2Cu -> 2Cu2+ + 4e-

Reduction: HNO3 + 3H+ + 2e- -> NO + 2H2O

Finally, we can combine the balanced half-reactions to obtain the balanced equation for the overall reaction:

2Cu + 8HNO3 + 6H+ -> 2Cu(NO3)2 + 2NO + 4H2O

Therefore, the balanced equation based on half-reactions is 2Cu + 8HNO3 + 6H+ -> 2Cu(NO3)2 + 2NO + 4H2O.

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A radioactive isotope has a half life of 25 minutes. How many half lives have occurred after 175 minutes?
5
6
7
8
Will give Brainliest

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Answer:

7

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