Which properties most like belong to a metal
1 answer:
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Answer:
Part A
The volume of the gaseous product is
Part B
The volume of the the engine’s gaseous exhaust is
Explanation:
Part A
From the question we are told that
The temperature is
The pressure is
The of
The chemical equation for this combustion is
The number of moles of that reacted is mathematically represented as
The molar mass of is constant value which is
So
The gaseous product in the reaction is and water vapour
Now from the reaction
2 moles of will react with 25 moles of
to give (16 + 18) moles of
and
So
1 mole of will react with 12.5 moles of
to give 17 moles of
and
This implies that
0.8754 moles of will react with (12.5 * 0.8754 ) moles of
to give (17 * 0.8754) of
and
So the no of moles of gaseous product is
From the ideal gas law
making V the subject
Where R is the gas constant with a value
Substituting values
Part B
From the reaction the number of moles of oxygen that reacted is
The volume is
No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as
Substituting values
Answer : The correct answer for the Theoretical Yield is 48.93 g of product .
Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .
Following are the steps to find theoretical yield .
Step 1) : Write a balanced reaction between Al and Cl₂ .
2 Al + 3 Cl₂→ 2 AlCl₃
Step 2: To find amount of product (AlCl₃) formed by Al .
Following are the sub steps to calculate amount of AlCl₃ formed :
a) To calculate mole of Al :
Given : Mass of Al = 35.5 g
Mole can be calculate by following formula :
Mole = 1.32 mol
b) To find mole ratio of AlCl₃ : Al
Mole ratio is calculated from balanced reaction .
Mole of Al in balanced reaction = 2
Mole of AlCl₃ in balanced reaction = 2.
Hence mole ratio of AlC; l₃ : Al = 2:2
c) To find mole of AlCl₃ formed :
Mole of AlCl₃ = 1.32 mol
d) To find mass of AlCl₃
Molar mass of AlCl₃ =
Mass of AlCl3 can be calculated using mole formula as:
Multiplying both side by
Mass of AlCl₃ = 176.00 g
Hence mass of AlCl₃ produced by Al is 176.00 g
Step 3) To find mass of product (AlCl₃) formed by Cl₂ :
Same steps will be followed to calculate mass of AlCl₃
a) Find mole of Cl₂
Mole of Cl₂ = 0.55 mol
b) Mole ratio of Cl₂ : AlCl₃
Mole of Cl₂ in balanced reaction = 3
Mole of AlCl₃ in balanced reaction = 2
Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3
c) To find mole of AlCl₃
Mole of AlCl3 = 0.367 mol
d) To find mass of AlCl₃ :
Multiplying both side by
Mass of AlCl₃ = 48.93 g
Hence mass of AlCl₃ produced by Cl₂ = 48.93 g
Step 4) To identify limiting reagent and theoretical yield :
Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .
The product AlCl₃ formed by Al = 176.00 g
The product AlCl₃ formed by Cl₂ = 48.93 g
Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .