Answer:
Explanation:
1. Write the skeleton equation for the half-reaction
NO₃⁻ ⟶ N₂O
2. Balance all atoms other than H and O
2NO₃⁻ ⟶ N₂O
3. Balance O by adding H₂O molecules to the deficient side.
2NO₃⁻ ⟶ N₂O + 5H₂O
4. Balance H by adding H⁺ ions to the deficient side.
2NO₃⁻ + 10H⁺ ⟶ N₂O + 5H₂O
5. Balance charge by adding electrons to the deficient side.
2NO₃⁻ + 10H⁺ + 8e⁻ ⟶ N₂O + 5H₂O
The amount of charge required to reduce 2 mol of NO₃⁻ is 8 F
<u>Answer:</u> The percent yield of the water is 31.98 %
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of methane = 6.58 g
Molar mass of methane = 16 g/mol
Putting values in equation 1, we get:
Given mass of oxygen gas = 14.4 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
The chemical equation for the combustion of methane is:
By Stoichiometry of the reaction:
2 moles of oxygen gas reacts with 1 mole of methane
So, 0.45 moles of oxygen gas will react with = of methane
As, given amount of methane is more than the required amount. So, it is considered as an excess reagent.
Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction
2 moles of oxygen gas produces 2 moles of water
So, 0.45 moles of oxygen gas will produce = moles of water
Molar mass of water = 18 g/mol
Moles of water = 0.45 moles
Putting values in equation 1, we get:
Experimental yield of water = 2.59 g
Theoretical yield of water = 8.1 g
Putting values in above equation, we get:
Hence, the percent yield of the water is 31.98 %