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Dmitry_Shevchenko [17]
3 years ago
14

The combustion of methane(CH4) produces carbon dioxide and water. Assume that 2.0 mol of CH4 burned in the presence of excess ai

r. What is the percentage yield if in an experiment the reaction produces 87.0 g of CO2?
Chemistry
1 answer:
harkovskaia [24]3 years ago
3 0

Answer:

The answer to your question is 98.9 %

Explanation:

Data

moles of methane = CH₄ = 2.0

excess air

Percent yield = ?

mass of CO₂ = 87 g

- Balanced chemical reaction

                CH₄  +  2O₂   ⇒  CO₂  +  2H₂O

      Reactants     Elements       Products

             1                    C                   1

             4                   H                   2

             4                   O                   2

- Calculate the molar mass of CH₄

CH₄ = 12 + 4 = 16 g

- Convert the moles to mass

                     16 g of CH₄ ----------------- 1 mol

                       x                -----------------  2 moles

                       x = (2 x 16) / 1

                       x = 32 g of CH₄

-Calculate the theoretical formation of CO₂

                     16 g of CH₄ ----------------- 44 g of CO₂

                     32 g of CH₄ ----------------  x

                            x = (32 x 44) / 16

                            x = 88 g of CO₂

-Calculate the Percent yield

     Percent yield = Actual yield/Theoretical yield x 100

     Percent yield = 87/88 x 100

    Percent yield = 98.9 %

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Consider the unbalanced equation for the oxidation of butene. C4H8 + 6O2 Right arrow. CO2 + H2O For each molecule of C4H8 that r
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Answer:

4 molecules of carbon dioxide (CO₂) and 4 molecules of water (H₂O)

Explanation:

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To balance the equation, we have to write a coefficient of 4 for CO₂ and a coefficient of 4 for H₂O, as follows:

C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O

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Question 4 of 10
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The number of atoms present in 2.5 mole of triatomic gas is equivalent to ​
RUDIKE [14]

Answer: 1.41725 X 10^{24} atoms

Explanation:

One male of any gas containing no. of atoms$=6.023\times10^{23}$

2.5 moles of gas containing no. of atoms$=2.5\times6.023\times10^{23}$$=15.0575\times10^{23}$

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6 0
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Deep sea divers use a mixture of helium and oxygen to breathe. Assume that a diver is going to a depth of 150 feet where the tot
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Answer:

4.525% is the percentage by volume of oxygen in the gas mixture.

Explanation:

Total pressure of the mixture = p = 4.42 atm

Partial pressure of the oxygen = p_1=0.20 atm

Partial pressure of the helium = p_2

p_1=p\times \chi_1 (Dalton law of partial pressure)

0.20 atm=4.42 atm\times \chi_1

\chi_1=\frac{0.20 atm}{4.42 atm}=0.04525

\chi_2=1-\chi_1=1-0.04525=0.95475

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According Avogadro law:

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Volume occupied by oxygen gas  =V_1

Total moles of gases = n = 1 mol

Total Volume of the gases = V

\frac{n_1}{V_1}=\frac{n}{V}

\frac{V_1}{V}=\frac{n_1}{n}=\frac{0.04525 mol}{1 mol}

Percent by volume of oxygen in the gas mixture:

\frac{V_1}{V}\times 100=\frac{0.04525 mol}{1 mol}\times 100=4.525\%

6 0
3 years ago
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