88g ( so sorry if this isn’t correct )
Answer:
NO would form 65.7 g.
H₂O would form 59.13 g.
Explanation:
Given data:
Moles of NH₃ = 2.19
Moles of O₂ = 4.93
Mass of NO produced = ?
Mass of produced H₂O = ?
Solution:
First of all we will write the balance chemical equation,
4NH₃ + 5O₂ → 4NO + 6H₂O
Now we will compare the moles of NO and H₂O with ammonia from balanced chemical equation:
NH₃ : NO NH₃ : H₂O
4 : 4 4 : 6
2.19 : 2.19 2.19 : 6/4 × 2.19 = 3.285 mol
Now we will compare the moles of NO and H₂O with oxygen from balanced chemical equation:
O₂ : NO O₂ : H₂O
5 : 4 5 : 6
4.93 : 4/5×4.93 = 3.944 mol 4.93 : 6/5 × 4.93 = 5.916 mol
we can see that moles of water and nitrogen monoxide produced from the ammonia are less, so ammonia will be limiting reactant and will limit the product yield.
Mass of water = number of moles × molar mass
Mass of water = 3.285 mol × 18 g/mol
Mass of water = 59.13 g
Mass of nitrogen monoxide = number of moles × molar mass
Mass of nitrogen monoxide = 2.19 mol × 30 g/mol
Mass of nitrogen monoxide = 65.7 g
Answer:
The answer to your question is 7.4 moles of Aluminum
Explanation:
Data
moles of Al = ?
moles of Al₂O₃ = 3.7
Balanced chemical reaction
4 Al + 3 O₂ ⇒ 2 Al₂O₃
To solve this problem use proportions and cross multiplication. Use the coefficients of the balanced chemical equation.
4 moles of Aluminum ----------------- 2 moles of Al₂O₃
x ----------------- 3.7 moles of Al₂O₃
x = (3.7 x 4) / 2
x = 14.8 / 2
x = 7.4 moles of Aluminum
B
A qualitative observation describes the characteristics of a substance without quantifying them.
