Determine the mass of carbon monoxide that is produced when 45.6 g of methane CH4 react with 73.2 g of oxygen gas, O2. The products are CO and H2O. Please check my work:
CH4 + 3O2 ---> 2CO + 2H20
nCH4=2.84 mol nO2=4.58 mol
Limiting Reactant:
1 mol CH4/2.84 mol CH4 = 3 mol O2/X O2 mol
xO2=8.52 mol
nCO=8.52 mol
mCO=nXM
=238.65 grams
To determine the mass of carbon monoxide (CO) produced, you first need to find the limiting reactant. The limiting reactant is the reactant that is completely consumed, determining the maximum amount of product that can be formed.
From your balanced equation, the stoichiometric ratio between methane (CH4) and oxygen gas (O2) is 1:3. This means that for every 1 mole of methane, you need 3 moles of oxygen gas to react.
Calculating the number of moles of each reactant:
Mass of CH4 = 45.6 g
Molar mass of CH4 = 12.01 g/mol (carbon) + 4(1.01 g/mol) (hydrogen) = 16.05 g/mol
nCH4 = massCH4 / molar massCH4
nCH4 = 45.6 g / 16.05 g/mol
nCH4 = 2.840 mol (rounded to three decimal places)
Mass of O2 = 73.2 g
Molar mass of O2 = 2(16.00 g/mol) = 32.00 g/mol
nO2 = massO2 / molar massO2
nO2 = 73.2 g / 32.00 g/mol
nO2 = 2.289 mol (rounded to three decimal places)
From the stoichiometry of the balanced equation, you can see that 1 mole of CH4 reacts with 3 moles of O2 to produce 2 moles of CO. Thus, you can determine the number of moles of CO that can be produced from the given limiting reactant, which is oxygen gas.
nO2 = 2.289 mol
nCO = (2/3) * nO2
nCO = (2/3) * 2.289 mol
nCO = 1.526 mol (rounded to three decimal places)
Now, to calculate the mass of CO produced, you need to multiply the number of moles of CO by its molar mass.
Molar mass of CO = 12.01 g/mol (carbon) + 16.00 g/mol (oxygen) = 28.01 g/mol
mCO = nCO * molar massCO
mCO = 1.526 mol * 28.01 g/mol
mCO = 42.78 g (rounded to two decimal places)
Therefore, the mass of carbon monoxide (CO) produced when 45.6 g of methane and 73.2 g of oxygen gas react is approximately 42.78 g.