17.8 grams of CH4 reacts with 55.4 g of )2 to from CO2 and H20
what is the mass of H20 formed? and how much excess reactant is leftover?
so i think that O2 is the limiting reagent and CH4 is the excess reagent
help?
I picked this up at the original post. I was out of pocket about three hours or I would have answered sooner. I have also included what to do with the excess reagent part of the problem. If you still have questions, post a new question but copy all of the work that's been done.
To find the mass of H2O formed and the excess reactant, we need to calculate the moles of each reactant and determine which reactant is the limiting reagent.
Let's start by calculating the moles of CH4 and O2.
1. Calculate the moles of CH4:
Molar mass of CH4 (carbon dioxide) = 12.01 g/mol + 4(1.01 g/mol) = 16.05 g/mol
Moles of CH4 = mass of CH4 / molar mass of CH4 = 17.8 g / 16.05 g/mol ≈ 1.11 mol
2. Calculate the moles of O2:
Molar mass of O2 (oxygen gas) = 2(16.00 g/mol) = 32.00 g/mol
Moles of O2 = mass of O2 / molar mass of O2 = 55.4 g / 32.00 g/mol ≈ 1.73 mol
Now, we need to determine the limiting reagent.
3. Use the balanced chemical equation to find the mole ratio between CH4 and O2:
CH4 + 2O2 → CO2 + 2H2O
From the equation, we see that 1 mole of CH4 reacts with 2 moles of O2 to produce 2 moles of H2O.
4. Compare the moles of O2 and the moles of H2O that would be produced if O2 was the limiting reagent:
Moles of H2O if O2 was limiting = 2 × (moles of O2) = 2 × 1.73 mol = 3.46 mol
Since the stoichiometric ratio is 2:1 (H2O:O2), we can conclude that the moles of H2O formed cannot exceed 3.46 mol.
5. Calculate the actual moles of H2O formed using the limiting reactant:
Moles of H2O formed = 1 mole of CH4 × (2 moles of H2O / 1 mole of CH4) = 1.11 mol
Now, we can calculate the mass of H2O formed.
6. Calculate the mass of H2O formed:
Molar mass of H2O (water) = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Mass of H2O formed = moles of H2O formed × molar mass of H2O = 1.11 mol × 18.02 g/mol ≈ 19.99 g
Therefore, the mass of H2O formed is approximately 19.99 grams.
Finally, to determine the amount of excess reactant leftover:
7. Calculate the moles of O2 required to react with all the CH4:
Moles of O2 required = 1.11 mol × (2 moles of O2 / 1 mole of CH4) = 2.22 mol
8. Calculate the remaining moles of O2:
Remaining moles of O2 = moles of O2 - moles of O2 required = 1.73 mol - 2.22 mol = -0.49 mol
Since we obtained a negative value, it means that there is no excess reactant leftover. It indicates that O2 was the limiting reagent.
To summarize:
- The mass of H2O formed is approximately 19.99 grams.
- There is no excess reactant leftover; O2 is fully consumed.