2so2+o2-2so3 if 96g so2 is added to 2 moles of oxygen at STP calculate the volume of oxygen used
To calculate the volume of oxygen used in the reaction, we need to use the balanced chemical equation and the ideal gas law.
First, let's balance the chemical equation given:
2SO2 + O2 -> 2SO3
According to the balanced equation:
- 2 moles of SO2 react with 1 mole of O2 to produce 2 moles of SO3.
Given that 96 g of SO2 is added, we need to convert this mass to moles. The molar mass of SO2 is approximately 64 g/mol.
Moles of SO2 = Mass / Molar mass
Moles of SO2 = 96 g / 64 g/mol
Moles of SO2 = 1.5 mol
Since the stoichiometry in the balanced equation is 2:1 for SO2 to O2, we can see that we need half the amount of moles of O2 compared to SO2.
Moles of O2 = 1.5 mol / 2
Moles of O2 = 0.75 mol
Now, we can use the ideal gas law to calculate the volume of oxygen used.
The ideal gas law is given by the equation:
PV = nRT
Where:
- P represents pressure (which is assumed to be constant at STP, 1 atm)
- V represents volume
- n represents the number of moles of gas
- R represents the ideal gas constant (0.0821 L·atm/(mol·K))
- T represents temperature (which is assumed to be constant at STP, 273.15 K)
Rearranging the ideal gas law to solve for volume V:
V = (nRT) / P
Plugging in the values:
V = (0.75 mol)(0.0821 L·atm/(mol·K))(273.15 K) / (1 atm)
Calculating:
V = 16.07 L
Therefore, the volume of oxygen used in the reaction is approximately 16.07 liters.