If 8.00g of methane is burned with 6.00g of oxygen. How much methane,oxygen,carbon dioxide,and water remain after the reaction is complete in grams?
This is a limiting reagent problem. I know that because amounts are given for BOTH reactants.
CH4 + 2O2 ==> CO2 + 2H2O
Convert 8.00g CH4 to mols. mol = grams/molar mass = about 0.500 g but you should confirm that as well as all of the other numbers. I estimate as I go along.
Convert 6.00 g O2 to mols. 6/32 = about 0.19 mols.
Using the coefficients in the balanced equation, convert mols CH4 to grams CO2.
That is 0.500 mols CH4 x (1 mol CO2/1 mol CH4) = 0.500 x 1/1 = 0.500 mol CO2.
Do the same for mols O2 to mols CO2.
0.19 x (1 mol CO2/2 mols O2) = 0.09 mols CO2.
This gave us two answers for mols CO2 and obviously one of them is not right. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Therefore, 0.19 mol CO2 is formed and O2 is the limiting reagent.
Using that value, calculate mols H2O formed. Using that value calculate mols CH4 used. CH4 remaining = initial mols - mols used.
You can convert mols CH4, CO2, and H2O to grams by g = mols x molar mass. There is no oxygen left.
Post your work if you stuck.
To solve this problem, we need to balance the chemical equation for the combustion of methane:
CH4 + 2O2 -> CO2 + 2H2O
From the equation, we can see that one mole of methane (CH4) reacts with two moles of oxygen (O2) to produce one mole of carbon dioxide (CO2) and two moles of water (H2O).
Step 1: Calculate the number of moles for each substance given:
Methane (CH4):
8.00g / molar mass of CH4
The molar mass of CH4 is:
(1 carbon atom x atomic mass of carbon) + (4 hydrogen atoms x atomic mass of hydrogen)
Oxygen (O2):
6.00g / molar mass of O2
The molar mass of O2 is:
(2 oxygen atoms x atomic mass of oxygen)
Step 2: Determine the limiting reactant:
The limiting reactant is the reactant that is completely consumed in the reaction and determines the amount of product formed. To find this, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation.
From the balanced equation, we see that one mole of CH4 requires two moles of O2.
If the number of moles of O2 is less than half the number of moles of CH4, then O2 is the limiting reactant.
If the number of moles of O2 is greater than half the number of moles of CH4, then CH4 is the limiting reactant.
Step 3: Calculate the moles of each product formed:
If CH4 is the limiting reactant:
- The moles of carbon dioxide (CO2) formed will be equal to the moles of CH4.
- The moles of water (H2O) formed will be twice the moles of CH4.
If O2 is the limiting reactant:
- The moles of carbon dioxide (CO2) formed will be half the moles of O2, since two moles of O2 produce one mole of CO2.
- The moles of water (H2O) formed will be equal to the moles of O2, since two moles of O2 produce two moles of H2O.
Step 4: Calculate the mass of each product formed:
Multiply the moles of each product by their respective molar masses.
Step 5: Calculate the remaining mass of unreacted CH4 and O2:
Subtract the mass of each reactant consumed from their initial masses.
Following these steps, you should be able to determine the amounts of methane, oxygen, carbon dioxide, and water remaining after the reaction is complete, all in grams.