4nh3 +7o2 -->4no2+6h20. what mass of h20 forms when 10.0 g of nh3 reacts with 25.3g of o2
N is 14 g/mol
H is 1 g/mol
O is 16 g/mol
so
NH3 is 14 + 3 = 17 g/mol
O2 is 32 g/mpl
you need 7 mols of O2 for every 4 mols of NH3
10 g NH3 = 10 g ( 1 mol / 17 g) = 0.588 mols
so we need
0.588 mol *7/4 = 1.03 mols of O2
1.03 mols* 32 g/mol = 32.9 grams of O2
WHOOOPS we only have 25.3 grams of O2
so
O2 is limiting reagent
25.3 grams O2 * 1 mol/ 32 g = 0.79 mols O2 ----> Use that
for every 7 mols O2 we 6 mols H2O
so mols H2O = (6/7)* .79 = 0.677 mols H2O produced
H2O = 2 + 16 = 18 g/mol
0.677 *18 = 12.2 grams H2O
To find the mass of H2O formed, we need to determine the limiting reactant between NH3 and O2. The limiting reactant is the one that runs out first and determines the maximum amount of product that can be formed.
To determine the limiting reactant, we can use the concept of stoichiometry and compare the number of moles of each reactant involved in the balanced equation.
1. Calculate the moles of NH3:
- Moles of NH3 = Mass of NH3 / Molar mass of NH3
- Molar mass of NH3 = 14.01 g/mol + (3 * 1.01 g/mol) = 17.03 g/mol
- Moles of NH3 = 10.0 g / 17.03 g/mol
2. Calculate the moles of O2:
- Moles of O2 = Mass of O2 / Molar mass of O2
- Molar mass of O2 = 16.00 g/mol + (2 * 1.01 g/mol) = 32.00 g/mol
- Moles of O2 = 25.3 g / 32.00 g/mol
3. Determine the limiting reactant:
- To find the limiting reactant, compare the mole ratio of NH3 to O2 from the balanced equation.
- From the balanced equation, the mole ratio of NH3 to O2 is 4:7.
- Divide the moles of NH3 by 4 and the moles of O2 by 7. The smaller value will be the limiting reactant.
4. Calculate the moles of H2O formed using the limiting reactant:
- According to the balanced equation, for every 4 moles of NH3, 6 moles of H2O are produced.
- Multiply the moles of the limiting reactant by the stoichiometric ratio (6 moles of H2O / 4 moles of NH3) to find the moles of H2O formed.
5. Calculate the mass of H2O formed:
- Mass of H2O = Moles of H2O * Molar mass of H2O
- Molar mass of H2O = 16.00 g/mol + (2 * 1.01 g/mol) = 18.02 g/mol
- Mass of H2O = Moles of H2O * 18.02 g/mol
By following these steps, you'll be able to calculate the mass of H2O formed when 10.0g of NH3 reacts with 25.3g of O2.
To find the mass of H2O formed when NH3 reacts with O2, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in the reaction, thus limiting the amount of product that can be formed.
Let's start by calculating the number of moles for each reactant using their respective molar masses.
Molar Mass of NH3 (ammonia) = 14.01 g/mol + 3(1.01 g/mol) = 17.03 g/mol
Molar Mass of O2 (oxygen) = 2(16.00 g/mol) = 32.00 g/mol
Number of moles of NH3 = mass of NH3 / molar mass of NH3
= 10.0 g / 17.03 g/mol
≈ 0.587 moles
Number of moles of O2 = mass of O2 / molar mass of O2
= 25.3 g / 32.00 g/mol
≈ 0.791 moles
Now, let's calculate the mole ratio of NH3 and H2O in the balanced equation:
From the balanced equation: 4NH3 + 7O2 -> 4NO2 + 6H2O
The mole ratio of NH3 to H2O is 4:6 (or simplified to 2:3).
Since we have 0.587 moles of NH3, we can use this mole ratio to determine the maximum moles of H2O that can be formed:
Number of moles of H2O = 0.587 moles NH3 × (3 moles H2O / 2 moles NH3)
≈ 0.880 moles H2O
Now, to calculate the mass of H2O formed, we can use the moles-to-mass conversion:
Mass of H2O = number of moles of H2O × molar mass of H2O
= 0.880 moles × 18.02 g/mol (molar mass of H2O)
≈ 15.85 g
Therefore, approximately 15.85 grams of H2O will form when 10.0 grams of NH3 reacts with 25.3 grams of O2.