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:
wheres the map? I can't see it even tho im using my eye glasses
False the heat does not matter with the wind
Answer:
30.06904
Explanation:
C3H4: Mass % C = 36.033 x 100 / 40.0641 = 89
286.8g of carbon dioxide forms when 95.6g of propane burns. The total atomic mass of the propane is 44(g). The total mass of the carbon dioxide formed when 44g of propane is burnt is 132(g). 44 divided by 44 equals 1. 132 divided by 44 equals 3. Therefore the amount of propane to carbon dioxide is in a 1:3 ratio. Therefore, 95.6 multiplied by 3 gives you the amount of carbon dioxide formed.