4NH3(g) +3O2(g) =2N2(g) +6H2O(l)
131.0 g of NH3 is mixed with 221.2 g of O2 and allowed to react. The reaction is complete when the NH3 is consumed. 107.8 g of N2 is formed and 36.60 g O2 remain unreacted. What mass of H2O is produced?
There is either a typo or wrong post. You should have 36.26 g O2 remaining unreacted if all of the other numbers are right. Then grams on left = grams on right.
grams NH3 used + g O2 used = g N2 formed + g H2O formed
131.0................+ (221,2-36.26) = 107.8 + ..........g H2O
solve for g H2O. The answer is approx 200 g
You can do this another way.
mols NH3 = grams/molar mass = ?
Using the coefficients in the balanced equatiopn, convert mols NH3 to mols H2O.
Then convert mols H2O to gramss H2O by g H2O = mols H2O x molar mass H2O
Post your work if you run into trouble.
To find the mass of H2O produced, we need to determine the amount of NH3 consumed in the reaction and then use the stoichiometry of the balanced equation to find the corresponding mass of H2O.
Let's start by calculating the amount of NH3 consumed in the reaction:
1. Calculate the molar mass of NH3:
- Molar mass of N = 14.01 g/mol
- Molar mass of H = 1.01 g/mol
- Molar mass of NH3 = (14.01 g/mol) + 3 * (1.01 g/mol) = 17.03 g/mol
2. Convert the given mass of NH3 to moles:
- Moles of NH3 = Mass of NH3 / Molar mass of NH3
- Moles of NH3 = 131.0 g / 17.03 g/mol = 7.68 mol
Since the balanced equation shows that the stoichiometric ratio between NH3 and N2 is 4:2, we can use this ratio to find the amount of N2 formed.
3. Calculate the moles of N2 formed:
- Moles of N2 = (7.68 mol NH3) x (2 mol N2 / 4 mol NH3)
- Moles of N2 = 7.68 mol / 2 = 3.84 mol
Now, we can find the molar mass of N2 and use it to calculate the mass of N2 formed.
4. Calculate the molar mass of N2:
- Molar mass of N2 = (2 * 14.01 g/mol) = 28.02 g/mol
5. Calculate the mass of N2 formed:
- Mass of N2 = Moles of N2 x Molar mass of N2
- Mass of N2 = 3.84 mol x 28.02 g/mol = 107.84 g
Now, let's move on to finding the mass of H2O produced.
6. Calculate the mass of O2 consumed:
- Mass of O2 consumed = Mass of O2 initial - Mass of O2 remaining
- Mass of O2 consumed = 221.2 g - 36.60 g = 184.6 g
7. Determine the moles of O2 consumed using the stoichiometric ratio between NH3 and O2:
- Moles of O2 consumed = (7.68 mol NH3) x (3 mol O2 / 4 mol NH3)
- Moles of O2 consumed = 7.68 mol x 3 / 4 = 5.76 mol
Finally, we can now use the stoichiometry of the balanced equation to calculate the mass of H2O produced.
8. Calculate the moles of H2O produced using the stoichiometry of the balanced equation:
- Moles of H2O = (5.76 mol O2) x (6 mol H2O / 3 mol O2)
- Moles of H2O = 5.76 mol x 6 / 3 = 11.52 mol
9. Calculate the molar mass of H2O:
- Molar mass of H2O = 2 * (1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
10. Calculate the mass of H2O produced:
- Mass of H2O = Moles of H2O x Molar mass of H2O
- Mass of H2O = 11.52 mol x 18.02 g/mol = 207.70 g
Therefore, the mass of H2O produced in the reaction is 207.70 g.