Answer:
The total pressure after one half is 6.375 atm.
Explanation:
The initial pressure of product is increases while the pressure of reactant would decrease.
Balanced chemical equation:
2N₂O → 2N₂ + O₂
The pressure of N₂O is 5.10 atm. The change in pressure would be,
N₂O = -2x
N₂ = +2x
O₂ = +x
The total pressure will be
P(total) = P(N₂O) + P(N₂) + P(O₂)
P(total) = ( 5.10 - 2x) + (2x) + (x)
P(total) = 5.10 + x
After one half life:
P(N₂O) = 1/2(5.10) = 5.10 - 2x
x = 5.10 - 1/2(5.10) /2
x = 5.10 - 0.5 (5.10) /2
x = 5.10 - 2.55 / 2
x = 2.55 /2 = 1.275 atm
Thus the total pressure will be,
P(total) = 5.10 + x
P(total) = 5.10 + 1.275
P(total) = 6.375 atm
The Balanced Chemical Equation is as follow;
4 KO₂ + 2 CO₂ → 2 K₂CO₃ + 3 O₂
First find out the Limiting Reagent,
According to equation,
284 g (4 moles) KO₂ reacted with = 44.8 L (2 moles) of CO₂
So,
27.9 g of KO₂ will react with = X L of CO₂
Solving for X,
X = (44.8 L × 27.9 g) ÷ 284 g
X = 4.40 L of CO₂
Hence, to consume 27.9 g of KO₂ only 4.40 L CO₂ is required, while, we are provided with 29 L of CO₂, it means CO₂ is in excess and KO₂ is is limited amount, Therefore, KO₂ will control the yield of K₂CO₃. So,
According to eq.
284 g (4 moles) KO₂ formed = 138.2 g of K₂CO₃
So,
27.9 g of KO₂ will form = X g of K₂CO₃
Solving for X,
X = (138.2 g × 27.9 g) ÷ 284 g
X = 13.57 g of K₂CO₃
So, 13.57 g of K₂CO₃ formed is the theoretical yield.
%age Yield = 13.57 / 21.8 × 100
%age Yield = 62.24 %
The balanced chemical reaction would be as follows:
2H2O2 = 2H2O + O2
We are given the amount of the peroxide that decomposes. Using this as the starting point for the calculations, we can determine the amount of O2 produced. We do as follows:
14.3 mol H2O2 ( 1 mol O2 / 2 mol H2O2 ) = 7.15 mol O2 produced
