I have posted this once and still cannot get the correct answer.
Determine the mass of carbon monoxide that is produced when 45.6 of methane, CH4, react with 73.2 g of oxygen , O2. The products are carbon monoxide and water vapour.
This is a limiting reagent problem. You know that because BOTH reactants are given. Here is the system for working limiting reagent problems. Copy and memorize the steps.
1. Write a balanced equation.
2CH4 + 3O2 ==> 2CO + 4H2O
2a. convert 45.6 g CH4 to moles. moles = grams/molar mass.
45.6/16 = 2.85
2b. Do the same for moles O2. 73.2/32 = 2.2875
3a. Using the coefficients in the balanced equation, convert moles CH4 to moles CO.
2.85 mols CH4 x (2 moles CO/2 moles CH4) = 2 x (2/2) = 2x1= 2 moles CO.
3b. Do the same for moles O2.
2.2875 moles O2 x (2 moles CO/3 moles O2) = 2.2875 x (2/3) = 1.525 moles CO.
3c. Note that the answers for 3a and 3b don't give the same value for moles CO which means one of them is wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value.
4. Convert the value from 3c to grams.
g = moles x molar mass. 1.525 x 28 = 42.70 g CO which I would round to 42.7. You need to confirm all of this as I estimated the molar masses of the above.
To determine the mass of carbon monoxide produced in the reaction, we need to first write a balanced equation for the reaction between methane (CH4) and oxygen (O2).
The balanced equation for the reaction is:
CH4 + 2O2 -> CO + 2H2O
From the balanced equation, we can see that one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon monoxide and two molecules of water.
To find the mass of carbon monoxide produced, we need to use the concept of stoichiometry. Stoichiometry is the calculation of reactants and products in chemical reactions.
First, calculate the molar mass of methane (CH4):
C = 12.01 g/mol
H = 1.01 g/mol (4 H atoms in methane)
Molar mass of methane (CH4):
12.01 g/mol + (4 * 1.01 g/mol) = 16.05 g/mol
Next, calculate the moles of methane reactant using the given mass:
moles of methane = mass of methane / molar mass of methane
moles of methane = 45.6 g / 16.05 g/mol ≈ 2.84 mol
According to the balanced equation, 1 mole of methane reacts to produce 1 mole of carbon monoxide. Therefore, the moles of carbon monoxide produced will also be 2.84 mol.
Now, let's find the mass of carbon monoxide produced using the molar mass of carbon monoxide (CO):
C = 12.01 g/mol
O = 16.00 g/mol
Molar mass of carbon monoxide (CO):
12.01 g/mol + 16.00 g/mol = 28.01 g/mol
mass of carbon monoxide = moles of carbon monoxide * molar mass of carbon monoxide
mass of carbon monoxide = 2.84 mol * 28.01 g/mol ≈ 79.57 g
Therefore, approximately 79.57 grams of carbon monoxide will be produced when 45.6 grams of methane react with 73.2 grams of oxygen.