If 96 g SO2 is added to 2 moles of oxygen at STP calculate the total volume of gas in the container at the end of the reaction
2SO2 + O2 ==> 2SO3
This is a limiting reagent (LR) problem. The non LR is called the ER for excess reagent.
mols SO2 = grams/molar mass = 96/64 = 1.5
How much O2 will be required to use all of the SO2. That's
1.5 mols SO2 x (1 mol O2/2 mols SO2) = 1.5 x 1/2 = 0.75. Do you have that much O2. Yes so SO2 is the LR and O2 is the ER.
How much SO3 is formed. That's 1.5 mols SO2 x (2 mols SO3/2 mols SO2) = 1.5 mols SO3 formed.
How much O2 is used. That's above @ 0.75 moles.
So you have 2 moles O2 initially - 0.75 mols O2 used in the reaction to leave an excess of 1.25 mols O2 unused. Total gas then is moles SO3 formed (1.5) + mols O2 left unreacted (1.25) = total mols gas after the reaction.
Final volume = moles SO3 x 22.4 L/mol@STP = ?
2SO2(g)+O2(g)>2SO3(g)
RFM of SO2= 32+2×16=64
no of mole of SO2 =96/64=1.5moles
mole ratio of SO2 to SO3 is 2:2, thus no of moles of SO3 is 1.5moles
1 mole at STP=22.4Litres
1.5 moles=??
=1.5×22.4
=33.6 Litres
To calculate the total volume of gas in the container at the end of the reaction, we first need to determine the balanced chemical equation for the reaction between SO2 and oxygen.
The balanced chemical equation is:
2 SO2 + O2 -> 2 SO3
From the balanced equation, we can see that 2 moles of SO2 react with 1 mole of O2 to produce 2 moles of SO3.
Given that 96 g of SO2 is provided, we need to convert this to moles to determine how many moles of SO2 we have:
Molar mass of SO2 = 32.06 g/mol
Number of moles of SO2 = 96 g / (32.06 g/mol) = 2.9957 mol (rounded to 4 decimal places)
Since the reaction ratio is 2 moles of SO2 to 1 mole of O2, we can say that the number of moles of O2 consumed is half the number of moles of SO2:
Number of moles of O2 consumed = 2.9957 mol / 2 = 1.4978 mol (rounded to 4 decimal places)
Now, we need to determine the total number of moles of gas at the end of the reaction. Since the volume of gas is directly proportional to the number of moles, we can use the ideal gas law equation to calculate the volume:
PV = nRT
Where:
P is the pressure (which is assumed to be at STP, so 1 atm)
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature in Kelvin (273 K at STP)
Rearranging the equation to solve for V:
V = nRT / P
Substituting the values, we get:
V = (1.4978 mol)(0.0821 L·atm/mol·K)(273 K) / 1 atm = 33.11 L (rounded to 2 decimal places)
Therefore, the total volume of gas in the container at the end of the reaction is 33.11 L.
To calculate the total volume of gas in the container at the end of the reaction, we need to determine the number of moles of each gas involved in the reaction and use the ideal gas law to calculate the volume.
First, let's determine the number of moles of SO2 and O2 in the reaction. We are given that 96 g of SO2 is added, and the molar mass of SO2 is approximately 64 g/mol. We can calculate the number of moles of SO2 by dividing the mass by the molar mass:
moles of SO2 = mass of SO2 / molar mass of SO2
= 96 g / 64 g/mol
= 1.5 moles
Next, we are given that there are 2 moles of O2.
Now, let's calculate the total number of moles of gas in the reaction:
Total moles of gas = moles of SO2 + moles of O2
= 1.5 moles + 2 moles
= 3.5 moles
Finally, we can use the ideal gas law to calculate the volume of the gas at STP (Standard Temperature and Pressure, which is 0 degrees Celsius and 1 atmosphere):
PV = nRT
Where:
P = pressure (1 atm)
V = volume
n = number of moles (3.5 moles)
R = ideal gas constant (0.0821 L⋅atm/mol⋅K)
T = temperature in Kelvin (0 degrees Celsius = 273 K)
Rearranging the equation to solve for V:
V = (nRT) / P
= (3.5 moles)(0.0821 L⋅atm/mol⋅K)(273 K) / 1 atm
= 78.5715 L
So, the total volume of gas in the container at the end of the reaction is approximately 78.6 L.