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 carbon monoxide and water vapour
This is a limiting reagent problem. You know that because BOTH reactants are given with their masses.
1. Write a balanced equation.
2CH4 + 3O2 --> 2CO + 4H2O
2. Convert 45.6 g CH4 and 73.2 g O2 to moles. moles = grams/molar mass.
3a. Using the coefficients in the balanced equation, convert moles CH4 to moles CO.
3b. Same procedure convert moles O2 to moles CO.
3c. It is more than likely that the answers frmo 3a and 3b will not be the same and one of them is wrong. The correct answer is ALWAYS the smaller value and the reactant producing that value is the limiting reagent.
4. Convert moles from 3c to grams.
g = mole x molar mass.
To determine the mass of carbon monoxide (CO) produced, we need to use the stoichiometry of the balanced chemical equation between methane (CH4) and oxygen (O2) to calculate the moles of reactants and products first. Then, we can convert the moles to mass using the molar mass of carbon monoxide.
Let's start by writing the balanced chemical equation for the reaction:
CH4 + 2O2 → CO2 + 2H2O
From the balanced equation, we can see that:
1 mole of CH4 produces 1 mole of CO
Step 1: Calculate the moles of methane (CH4) and oxygen (O2):
Molar mass of CH4 (methane) = 12.01 g/mol (C) + 4 * 1.01 g/mol (H) = 16.04 g/mol
Number of moles of CH4 = 45.6 g / 16.04 g/mol = 2.84 mol (rounded to two decimal places)
Molar mass of O2 (oxygen) = 2 * 16.00 g/mol = 32.00 g/mol
Number of moles of O2 = 73.2 g / 32.00 g/mol = 2.29 mol (rounded to two decimal places)
Step 2: Determine the limiting reactant:
To find out which reactant is limiting, we compare the mole ratio between CH4 and O2 in the balanced equation:
From the balanced equation: 1 mole of CH4 requires 2 moles of O2
Using the number of moles calculated above, we find the mole ratio:
CH4 : O2 = 2.84 mol : 2.29 mol
The ratio is approximately 1.24 : 1, indicating that O2 is the limiting reactant. So, all of the O2 will be consumed, and the amount of CH4 will be in excess.
Step 3: Calculate the moles of CO formed:
Since O2 is the limiting reactant and 1 mole of CH4 produces 1 mole of CO, the number of moles of CO formed will be equal to the number of moles of O2.
Number of moles of CO formed = 2.29 mol (the number of moles of O2)
Step 4: Convert moles of CO to mass:
Molar mass of CO = 12.01 g/mol (C) + 16.00 g/mol (O) = 28.01 g/mol
Mass of CO formed = Number of moles of CO * Molar mass of CO
= 2.29 mol * 28.01 g/mol
= 64.1 g (rounded to one decimal place)
Therefore, the mass of carbon monoxide (CO) produced when 45.6 g of methane (CH4) reacts with 73.2 g of oxygen gas (O2) is approximately 64.1 g.