Respuesta :
Answer:
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 %
Answer:
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 %
