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emmainna [20.7K]
3 years ago
13

Gaseous methane (CH4) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 28

.2 g of carbon dioxide is produced from the reaction of 15.1 g of methane and 81.2 g of oxygen gas, Calculate the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it.
Chemistry
1 answer:
elena-14-01-66 [18.8K]3 years ago
6 0

Answer:

68.1% is percent yield of the reaction

Explanation:

The reaction of methane with oxygen is:

CH₄ + 2O₂ → CO₂ + 2H₂O

<em>Where 2 moles of oxygen react per mole of CH₄</em>

<em />

Percent yield is:

Actual yield (28.2g CO₂) / Theoretical yield * 100

To solve this question we need to find theoretical yield finding limiting reactant :

<em>Moles CH₄:</em>

15.1g CH₄ * (1mol / 16.04g) = 0.9414 moles

<em>Moles O₂:</em>

81.2g * (1mol / 32g) = 2.54 moles

For a complete reaction of 0.9414 moles of CH₄ are needed:

0.9414 moles CH₄ * (2 mol O₂ / 1mol CH₄) = 1.88 moles of O₂. As there are 2.54 moles, O₂ is in excess and <em>CH₄ is limiting reactant</em>

In theoretical yield, the moles of methane added = Moles of CO₂ produced. That is 0.9414 moles CO₂. In grams = Theoretical yield:

0.9414 moles CO₂ * (44.01g / mol) = 41.43g CO₂

Percent yield: 28.2g CO₂ / 41.43g CO₂ * 100=

<h3>68.1% is percent yield of the reaction</h3>
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3 years ago
“Dry ice” is the solid form of carbon dioxide. Determine the number of mass of CO2 gas in grams that are present in 61.8 L of CO
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Explanation:

Given data:

Mass in gram of CO₂ = ?

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PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

1 atm × 61.8 L = n ×0.0821 atm.L/ mol.K   × 273.15 k

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