For
the reaction 2SO2 + O2 -> 2 SO3, we first determine which is the excess
reactant between SO2 and O2. We list down the molar mass of the reactants:
Molar
mass of SO2 = 64.0638 g/mol
Molar
mass of O2 = 32 g/mol
Using
the stoichiometry of the reaction, we then calculate the amount of oxygen that
will react with 90.0 g of SO2.
90.0
g SO2 x 1 mol SO2/64.0638 g x 1 mol O2/ 2 mol SO2 x 32 g O2/mol = 22.4776 g O2
<span>
</span>
<span>Thus,
we can conclude that O2 is the excess reactant while SO2 is the limiting
reactant. Subtracting 22.4776 g O2 from the initial 100.0 g O2, we get 77.5224
g O2 left after the complete reaction of 90.0 g SO2. </span>
Answer: The atom has three components the electrons, neutrons and protons.
Explanation:
J.J Thomson is responsible for the discovery of electron. He discovered the electrons while determining the properties of cathode rays in 1897. Rutherford is responsible for the discovery of proton during 1909 while performing the gold foil experiment. W. Bothe and H. Becker is credited to the discovery of neutrons.
Explanation:
At sea level, the atmospheric pressure would be a little over 100 kPa (one atmosphere or 760 mm Hg). If we climb to the top of Mount Everest (the highest mountain in the world at 29,029 feet or 8848 meters), the atmospheric pressure will drop to slightly over 30 kPa (about 0.30 atmospheres or 228 mm Hg).
Hope it is helpful for you
Answer:
66 g of CO₂
Solution:
The Balance Chemical Reaction is as follow,
C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O
Or,
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O ------- (1)
Step 1: Find out the limiting reagent as;
According to Equation 1,
56.1 g (2 mole) C₂H₂ reacts with = 160 g (5 moles) of O₂
So,
125 g of C₂H₂ will react with = X g of O₂
Solving for X,
X = (125 g × 160 g) ÷ 56.1 g
X = 356.5 g of O₂
It means for total combustion of Ethylene we require 356.5 g of O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.
Step 2: Calculate Amount of CO₂ produced as;
According to Equation 1,
160 g (5 mole) O₂ produces = 176 g (4 moles) of CO₂
So,
60.0 g of O₂ will produce = X g of CO₂
Solving for X,
X = (60.0 g × 176 g) ÷ 160 g
X = 66 g of CO₂