45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)
The balanced equation for the reaction is given below:
2Al + 3Cl₂ —> 2AlCl₃
From the balanced equation above,
2 atoms of Al required 3 molecules of Cl₂.
With the above information, we can determine the number of molecules of Cl₂ needed to react with 30 atoms of Al. This can be obtained as follow:
From the balanced equation above,
2 atoms of Al required 3 molecules of Cl₂.
Therefore,
30 atoms of Al will require = 
= 45 molecules of Cl₂.
Thus, 45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)
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382.85 Celsius is the temperature does 0.750 moles of an ideal gas occupy a volume of 35.9 L at 114 kPa.
Explanation:
Given data:
number of moles of the gas = 0.75 moles
volume of the gas = 35.9 liters
pressure of the gas = 114 KPa or 1.125 atm
R = 0.0821 latm/moleK
temperature of the gas T = ?
The equation used to calculate temperature from above data is ideal gas law equation.
the equation is :
PV = nRT
T = 
Putting the values in the above rewritten equation:
T = 
T = 655.9 K
To convert kelvin into celsius, formula used is
K = 273.15+ C
putting the values in the equation
C = 656 - 273.15
= 382.85 Celsius
( The rapid motion and collisions of molecules ) with the walls of a container causes pressure.
Source; https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Pressure
Answer:
The percent yield of chloro-ethane in the reaction is 82.98%.
Explanation:
Moles of ethane = 
Moles of chlorine gases =
As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.
This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.
According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.
Then 9.1549 moles of chlorien gas will give:
of chloro-ethane
Mass of 9.1549 moles of chloro-ethane:
9.1549 mol × 64.5 g/mol = 590.4910 g
Theoretical yield of chloro-ethane: 590.4910 g
Given experimental yield of chloro-ethane: 490.0 g
The percent yield of chloro-ethane in the reaction is 82.98%.
