A 8.6 g sample of methane and 15.6 g sample of oxygen react according to the reaction in the video. identify the limiting reactant and calculate the mass of carbon dioxide that could be formed. 22.8 g carbon dioxide comes from 8.6 g of ch4 or 18.7 g carbon dioxide comes from 15.6 g o that means the 15.6 g of oxygen is still the limiting reactant because it gets used up and only makes 10.7 g of co2. 23.6 g carbon dioxide comes from 8.6 g of ch4 or 10.7 g carbon dioxide comes from 15.6 g o that means the 15.6 g of oxygen is still the limiting reactant because it gets used up and only makes 10.7 g of co2. 10.7 g carbon dioxide comes from 8.6 g of ch4 or 23.6 g carbon dioxide comes from 15.6 g o that means the 8.6 g of co2 is still the limiting reactant because it gets used up and only makes 10.7 g of co2. submit rewatch

Question

Bishop

Chemistry

14 July, 20:45

A 8.6 g sample of methane and 15.6 g sample of oxygen react according to the reaction in the video. identify the limiting reactant and calculate the mass of carbon dioxide that could be formed. 22.8 g carbon dioxide comes from 8.6 g of ch4 or 18.7 g carbon dioxide comes from 15.6 g o that means the 15.6 g of oxygen is still the limiting reactant because it gets used up and only makes 10.7 g of co2. 23.6 g carbon dioxide comes from 8.6 g of ch4 or 10.7 g carbon dioxide comes from 15.6 g o that means the 15.6 g of oxygen is still the limiting reactant because it gets used up and only makes 10.7 g of co2. 10.7 g carbon dioxide comes from 8.6 g of ch4 or 23.6 g carbon dioxide comes from 15.6 g o that means the 8.6 g of co2 is still the limiting reactant because it gets used up and only makes 10.7 g of co2. submit rewatch

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Answers (1)

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Randy Kane · Answer 1

23.6 g carbon dioxide comes from 8.6 g of CH4 or 10.7 g carbon dioxide comes from 15.6 g O that means the 15.6 g of oxygen is still the limiting reactant because it gets used up and only makes 10.7 g of CO2.

Explanation:

1) Balanced chemical equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

2) mole ratios:

1 mol CH₄ : 2mol O₂ : 1 mol CO₂ : 2 mol H₂O

3) molar masses

CH₄: 16.04 g/mol

O₂: 32.0 g/mol

CO₂: 44.01 g/mol

4) Convert the reactant masses to number of moles, using the formula

number of moles = mass in grams / molar mass

CH₄: 8.6g / 16.04 g/mol = 0.5362 moles

O₂: 15.6 g / 32.0 g/mol = 0.4875 moles

5) If the whole 0.5632 moles of CH₄ reacted that yields to the same number of moles of CO₂ and that is a mass of:

mass of CO₂ = number of moles x molar mass = 23.60 g of CO₂

Which is what the first part of the answer says.

6) If the whole 0.4875 moles of O₂ reacted that would yield 0.4875 / 2 = 0.24375 moles of CO₂, and that is a mass of:

mass of CO₂ = 0.4875 grams x 44.01 g/mol = 10.7 grams of CO₂.

Which is what the second part of the answer says.

7) From the mole ratio you know infere that 0.5362 moles of CH₄ needs more twice number of moles of O₂, that is 1.0724 moles of O₂, and since there are only 0.4875 moles of O₂, this is the limiting reactant.

Which is what the chosen answer says.

8) From the mole ratios 0.4875 moles of O₂ produce 0.4875 / 2 moles of CO₂, and that is:

0.4875 / 2 mols x 44.01 g/mol = 10.7 g of CO₂, which is the last part of the answer.