Answer:
Iron(III) oxide and its common name is Ferric Oxide
Explanation:
CH4 : H2O
1 : 2
number of moles of H2O = 1.00 x 2
number of moles of H2O = 2.00mol
mass = number of moles x molar mass
mass of H2O = 2.00 x (1 + 1 + 16)
mass of H2O = 36g
Answer:
Theoretical yield = 2.5 g
Explanation:
Given data:
Mass of sodium = 79.7 g
Mass of water = 45.3 g
Theoretical yield of hydrogen gas = ?
Solution:
Chemical equation:
2Na + 2H₂O → 2NaOH + H₂
Number of moles of sodium:
Number of moles = mass/ molar mass
Number of moles = 79.7 g / 23 g/mol
Number of moles = 3.5 mol
Number of moles of water:
Number of moles = mass/ molar mass
Number of moles = 45.3 g / 18g/mol
Number of moles = 2.5 mol
Now we will compare the moles of hydrogen gas with water and sodium.
H₂O : H₂
2 : 1
2.5 : 1/2×2.5 =1.25 mol
Na : H₂
2 : 1
3.5 : 1/2×3.5 =1.75 mol
water will be limiting reactant.
Theoretical yield:
Mass = number of moles × molar mass
Mass = 1.25 mol × 2 g/mol
Mass = 2.5 g
Answer:
3.64g
Explanation:
Given parameters:
Mass of NH₃ = 18.1g
Mass of Cu₂O = 90.4g
Unknown:
Limiting reactant = ?
Mass of N₂ formed = ?
Solution:
The reaction equation is given as:
Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O
The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;
Number of moles =
Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol
Molar mass of NH₃ = 14 + 3(1) = 17g/mol
Number of moles of Cu₂O =
= 0.13moles
Number of moles of NH₃ =
= 5.32moles
From this reaction;
1 mole of Cu₂O combines with 2 mole of NH₃
So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃
= 0.26moles of NH₃
Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;
Mass of N₂;
Mass = number of moles x molar mass
1 mole of Cu₂O will produce 1 mole of N₂
0.13 mole of Cu₂O will produce 0.13 mole of N₂
Mass = 0.13 x (2 x 14) = 3.64g