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When 16 g of methane (CH4) and 32 g of oxygen (O2) reacted to produce carbon dioxide and water, 11 g of carbon dioxide was produced. Calculate the percent yield of carbon dioxide in this reaction. A) 5.0% B) 10% C) 25% D) 50%

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Neetoo

Answer:

Percent yield = 50%

Explanation:

Given data:

Mass of CHâ‚„ = 16 g

Mass of Oâ‚‚ = 32 g

Mass of COâ‚‚ = 11 g

Percent yield of COâ‚‚ = ?

Solution:

Chemical equation:

CH₄ + 2O₂    →  CO₂ + 2H₂O

Number of moles of CHâ‚„:

Number of moles = mass/ molar mass

Number of moles = 16 g /16 g/mol

Number of moles = 1 mol

Number of moles of Oâ‚‚:

Number of moles = mass/ molar mass

Number of moles = 32 g /32 g/mol

Number of moles = 1 mol

Now we will compare the moles of COâ‚‚ with both reactant.

                             O₂             :            CO₂

                              2              :               1

                              1               :          1/2×1= 0.5 mol

                          CH₄              :            CO₂

                             1               :               1

Number of moles of COâ‚‚ produced by oxygen are less so it will limiting reactant.

Theoretical yield:

Mass of COâ‚‚:

Mass = number of moles × molar mass

Mass = 0.5 mol × 44 g/mol

Mass = 22 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 11 g/ 22 g × 100

Percent yield = 50%

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