Match each process to where it occurs in the carbon cycle
1 answer:
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Answer:
36.37% is the percent yield of the reaction.
Explanation:
1)0.650 L nitrogen gas , at 295 K and 1.01 bar.
Let the moles of nitrogen gas be n.
Pressure of the gas ,P= 1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)
Temperature of the gas = T = 295 K
Volume of the gas = V = 0.650 L
Using an ideal gas equation:
2) Moles of ammonia gas=
Moles of oxygen gas =
According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.
Then,0.1101 mol of oxygen will react with:
of ammonia.
Hence, oxygen gas is in limiting amount and act as limiting reagent.
3) Theoretical yield of nitrogen gas :
According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.
Then 0.1101 mol of oxygen will give:
of nitrogen.
Theoretical yield of nitrogen gas = 0.0734 mol
Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol
Percentage yield:
Percentage yield of the reaction:
36.37% is the percent yield of the reaction.
Answer:
1,812 wt%
Explanation:
The reactions for this titration are:
2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻
I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻
The moles in the end point of S₂O₃⁻ are:
0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:
4,463x10⁻⁴ moles of S₂O₃⁻× = 2,2315x10⁻⁴ moles of I₃⁻
As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:
2,2315x10⁻⁴ moles of I₃⁻× = 4,463x10⁻⁴ moles of Ce(IV). These moles are:
4,463x10⁻⁴ moles of Ce(IV)× = <em>0,0625 g of Ce(IV)</em>
As the sample has a 3,452g, the weight percent is:
0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>
I hope it helps!