Gaseous methane (ch4) reacts with gaseous oxygen gas (o2) to produce gaseous carbon dioxide (co2) and gaseous water (h2o) if 0.391 g of carbon dioxide is produced from the reaction of 0.16 g of methane and 0.84 g of oxygen gas, calculate the percent yield of carbon dioxide.

The % yied of carbon dioxide = 89.1 %

Explanation:

Step 1: Data given

Mass of CO2 = 0.391 grams

Mass of methane = 0.16 grams

Mass of oxygen gas = 0.84 grams

Molar mass of CO2 = 44.01 g/mol

Molar mass of methane (CH4) = 16.04 g/mol

Molar mass of oxygen gas (O2) = 32 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 3: Calculate moles of CH4

Moles CH4 = mass CH4 / Molar mass CH4

Moles CH4 = 0.16 grams / 16.04 g/mol

Moles CH4 = 0.00998 mol

Step 4: Calculate moles of O2

Moles O2 = mass O2 / Molar mass O2

Moles O2 = 0.84 grams / 32 g/mol

Moles O2 =0.02625 mol

Step 5: Calculate limiting reactant

For 1 mole CH4 consumed, we need 2 moles O2, to produce 1 mole of CO2

CH4 is the limiting reactant. It will completely be consumed (0.00998 moles)

O2 is in excess. There will be consumed 2*0.00998 = 0.01996 mol

There will remain 0.02625 - 0.01996 = 0.00629 mol O2

Step 6: Calculate moles carbon dioxide

For 1 mole CH4 consumed, we need 2 moles O2, to produce 1 mole of CO2

For 0.00998 moles of CH4, we have 0.00998 moles of CO2

Step 7: Calculate theoretical mass of CO2

Mass of CO2 = moles of CO2 *Molar mass of CO2

Mass of CO2 =0.00998 moles * 44.01 g/mol

Mass of CO2 = 0.439 grams

Step 8: Calculate the % tield of CO2

% yield = actual yield / theoretical yield

% yield = (0.391 grams / 0.439 grams) *100%

% yield = 89.1%

The % yied of carbon dioxide = 89.1 %

percent yield of carbon dioxide is 46.5 %

Explanation:

Formula for the above reaction is  

CH₄+2O₂→CO₂+2H₂OCH₄+2O₂→CO₂+2H₂O

Firstly I will calculate the moles of reactants

Moles of CH₄= given mass/ Molar mass= 0.16g /16gmole=0.01 moles of CH₄(limiting reactant)

Moles of Oxygen O₂= given mass/ Molar mass = 0.84 g/32gmole=0.026 moles of O₂ (reactant in excess)

Thus, moles  of CO₂ = 0.01  

Mass of CO₂= 0.01 mole × 44 g mole=0. 44 grams

Therefore,

For percent yield=experimental yield /theoretical yield×100%.

% yield= 0.391g/0.84×100%=46.5 %

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