Gaseous methane
CH4
will react with gaseous oxygen
O2
to produce gaseous carbon dioxide
CO2
and gaseous water
H2O
. Suppose 0.481 g of methane is mixed with 2.9 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.
You do the math and s.f.
This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH of the reactants.
1.Write and balance the equation.
CH4 + 2O2 ==> CO2 + 2H2O
2a. mols CH4 = grams/molar mass = ?
2b. mols O2 = grams/molar mass = ?
3a.
Using the coefficients in the balanced equation, convert mols CH4 in 2a to mols CO2.
3b. Do the same and convert mols O2 in step 2b to mols CO2
3c. It is likely the two values in 3a and 3b will not agree which means one of them is not right.
3d. The correct answer in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
4. Now convert the smaller value into grams CO2. g = mols x molar mass
Print this up. Use it for all of your LR problems.
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To calculate the maximum mass of carbon dioxide (CO2) produced, we need to determine the limiting reactant in the reaction. The limiting reactant is the one that is entirely consumed and determines the maximum amount of product formed.
1. Calculate the number of moles of methane (CH4) and oxygen (O2) present:
Moles of CH4 = mass / molar mass = 0.481 g / 16.04 g/mol = 0.030 moles
Moles of O2 = mass / molar mass = 2.9 g / 32.00 g/mol = 0.091 moles
2. Use the mole ratio between CH4 and O2 from the balanced chemical equation to determine the stoichiometric ratio:
CH4 + 2O2 -> CO2 + 2H2O
Moles of CH4 : Moles of O2 = 1 : 2
Since the stoichiometric ratio is 1:2, it means that for every 1 mole of CH4, we need 2 moles of O2.
3. Determine the limiting reactant:
The limiting reactant will be the one that has fewer moles. In this case, it is CH4 with 0.030 moles.
4. Calculate the moles of CO2 formed:
From the balanced chemical equation, we can see that for every mole of CH4 reacted, 1 mole of CO2 is produced.
Moles of CO2 = 0.030 moles of CH4 * (1 mole of CO2 / 1 mole of CH4) = 0.030 moles
5. Calculate the mass of CO2 produced:
Mass = moles * molar mass = 0.030 moles * 44.01 g/mol = 1.320 g
Therefore, the maximum mass of carbon dioxide that could be produced in this reaction is 1.320 grams.
To calculate the maximum mass of carbon dioxide produced, you first need to determine the limiting reagent among the reactants.
1. Start by determining the number of moles for each reactant.
- The molar mass of methane (CH4) is 16.04 g/mol (4 * 12.01 + 1 * 4.01).
- Convert the given mass of methane (0.481 g) to moles by dividing it by the molar mass:
moles of methane = 0.481 g / 16.04 g/mol = 0.03 mol.
- The molar mass of oxygen (O2) is 32.00 g/mol (2 * 16.00).
- Convert the given mass of oxygen (2.9 g) to moles by dividing it by the molar mass:
moles of oxygen = 2.9 g / 32.00 g/mol = 0.091 mol.
2. Determine the stoichiometric ratio between methane and carbon dioxide in the balanced equation. The balanced equation shows that for every 1 mole of methane, 1 mole of carbon dioxide is produced.
3. Compare the moles of methane and oxygen to determine the limiting reagent. The reactant with the lesser number of moles is the limiting reagent.
- In this case, methane has 0.03 mol and oxygen has 0.091 mol. Methane has lesser moles.
4. Calculate the amount of carbon dioxide produced using the stoichiometric ratio.
- From the stoichiometric ratio, 1 mole of methane produces 1 mole of carbon dioxide.
- So, 0.03 mol of methane would produce 0.03 mol of carbon dioxide.
5. Convert the moles of carbon dioxide to grams using its molar mass.
- The molar mass of carbon dioxide (CO2) is 44.01 g/mol (12.01 * 1 + 16.00 * 2).
- Multiply the moles of carbon dioxide (0.03 mol) by its molar mass:
mass of carbon dioxide = 0.03 mol * 44.01 g/mol = 1.32 g.
Therefore, the maximum mass of carbon dioxide that could be produced by the reaction is 1.32 grams.