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 %
<span>Na --> Na+ + e- E = +2.71
2H2O + 2 e- --> H2 + 2 OH- E = -0.83
Delta E = 2.71 - 0.83 = 1.88 Volts
You know the reaction is spontaneous because Ecell is positive.
Delta Go = - n F Delta E
Delta Go = - 2mol (96485 J/volt mol) (1.88 V)
Delta go = -3.62X10^5 J/mol = -326 kJ/mol
Yes, it is VERY spontaneous.</span>
Answer:
They indicate the number of each chemical species that reacts or is formed. Methane and oxygen (oxygen is a diatomic — two-atom — element) are the reactants, while carbon dioxide and water are the products. All the reactants and products are gases (indicated by the g's in parentheses).
Answer: 46.6 grams
Explanation:
Given : density of hexane = 0.660 g/ml
volume of hexane = 20 ml
Mass of hexane =
moles of hexane =
According to stoichiometry :
2 moles of hexane reacts with 19 moles of oxygen
Thus 0.14 moles hexane will react with = moles of oxygen
Mass of oxygen =
Thus minimum mass of oxygen required for the complete reaction of 20.0 mL of hexane is 46.6 grams.