What volume of oxygen is needed to react with solid sulfur to form 6.20 L of sulfur dioxide?
Write the equation and balance it.
S + O2 ==> SO2 (Or your teacher may prefer that you write it as
S8 + O2 ==> SO2.
Convert 6.20 L SO2 to moles. 6.20 x (1 mol/22.4 L) = ??
Using the coefficients in the balanced equation, convert mols SO2 to moles O2.
Now convert moles O2 to liters. L = moles x 22.4 L/mol = ??
the answer is 6.20 L
its 6.20 L
To determine the volume of oxygen needed to react with solid sulfur, we need to use the balanced chemical equation for the reaction between sulfur and oxygen to form sulfur dioxide:
2S (s) + 3O2 (g) → 2SO2 (g)
From the balanced equation, we can see that two moles of solid sulfur (S) react with three moles of oxygen gas (O2) to produce two moles of sulfur dioxide (SO2).
Now let's proceed with the calculations:
Step 1: Convert the given volume of sulfur dioxide to moles
We are given that the volume of sulfur dioxide (SO2) formed is 6.20 L. To convert this volume to the number of moles, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (assumed to be constant)
V = volume of gas
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (assumed to be constant)
Since the pressure, temperature, and volume are constant, we can simplify the equation to:
n = (PV) / RT
Assuming standard temperature and pressure (STP), we use the values:
P = 1 atm
V = 6.20 L
T = 273 K
This gives us:
n = (1 atm * 6.20 L) / (0.0821 L·atm/mol·K * 273 K)
n ≈ 0.271 mol
Therefore, we have approximately 0.271 moles of sulfur dioxide (SO2).
Step 2: Use the mole ratio to determine the moles of oxygen
From the balanced chemical equation, we know that 2 moles of sulfur dioxide (SO2) are produced from 3 moles of oxygen gas (O2):
2 mol SO2 : 3 mol O2
By setting up a proportion, we can determine the moles of oxygen gas needed:
0.271 mol SO2 : x
Using the proportion:
0.271 mol SO2 / 2 mol SO2 = x / 3 mol O2
Simplifying this equation:
x = (3 mol O2 * 0.271 mol SO2) / 2 mol SO2
x ≈ 0.406 mol O2
Therefore, we need approximately 0.406 moles of oxygen gas (O2).
Step 3: Convert the moles of oxygen to volume
Since we have the number of moles of oxygen gas, we can convert it to volume using the ideal gas law equation:
PV = nRT
Assuming standard temperature and pressure (STP), we use the values:
P = 1 atm
V = volume of gas (unknown)
T = 273 K
n = 0.406 mol
Solving for V:
V = (nRT) / P
V = (0.406 mol * 0.0821 L·atm/mol·K * 273 K) / 1 atm
V ≈ 9.30 L
Therefore, approximately 9.30 L of oxygen gas (O2) is needed to react with solid sulfur to form 6.20 L of sulfur dioxide (SO2).