2 SO2 (g) + O2 (g) → 2 SO3↑
Assuming STP, how many grams of oxygen are needed to produce 19.8 liters of SO3?
When gases are involved you may use a shortcut; i.e., use volumes as if volume = mols.
19.8 L SO3 x (1 mol O2/2 mols SO3) = ? L O2.
Then grams = mols x molar mass.
02+2SO2=2SO3
volume(O2)=19.8L*(1LO2/2LSO3)
=9.9L of O2
To determine the grams of oxygen needed to produce 19.8 liters of SO3 at STP, we can use stoichiometry.
First, we need to find the number of moles of SO3 produced. Since 1 mole of gas at STP occupies 22.4 liters, we can calculate the moles of SO3 as follows:
19.8 liters SO3 x (1 mole SO3 / 22.4 liters) = 0.884 moles SO3
According to the balanced equation, the molar ratio between SO3 and O2 is 2:1. This means that for every 2 moles of SO3 produced, 1 mole of O2 is required.
So, we can calculate the moles of O2 needed as follows:
0.884 moles SO3 x (1 mole O2 / 2 moles SO3) = 0.442 moles O2
Now, we can use the molar mass of O2 to convert moles to grams. The molar mass of O2 is 32 g/mol.
0.442 moles O2 x (32 g / 1 mole) = 14.144 grams of O2
Therefore, 19.8 liters of SO3 at STP requires 14.144 grams of O2.
To find out how many grams of oxygen are needed to produce 19.8 liters of SO3, we can use the given balanced chemical equation:
2 SO2 (g) + O2 (g) → 2 SO3 (g)
From the equation, we can see that the ratio of oxygen to sulfur dioxide is 1:2. This means that for every 2 moles of sulfur dioxide (SO2) reacted, we need 1 mole of oxygen gas (O2) to react.
First, we need to convert the given volume of SO3 gas to the number of moles. At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies 22.4 liters.
Given: Volume of SO3 = 19.8 liters
Moles of SO3 = Volume of SO3 / Molar volume at STP
= 19.8 L / 22.4 L/mol
= 0.89 moles
Since the balanced equation shows a 1:1 ratio between oxygen and sulfur dioxide, we can conclude that 0.89 moles of SO3 requires 0.89 moles of O2.
To find the mass of oxygen, we need to multiply the moles of oxygen by its molar mass. The molar mass of oxygen (O2) is approximately 32.00 g/mol.
Mass of O2 = Moles of O2 * Molar mass of O2
= 0.89 moles * 32.00 g/mol
= 28.48 grams
Therefore, approximately 28.48 grams of oxygen are needed to produce 19.8 liters of SO3 at STP.