CS2+3O2 --> CO2+2SO2
What is the volume of sulfur dioxide, in milliliters,produced when 27.9 mL O2 reacts with carbon disulfide
For gas problems like this you may take a shortcut and use volume as mols.
27.3 mL O2 x (2 mols SO2/3 mols O2) = ? mL SO2.
To find the volume of sulfur dioxide (SO2) produced, we first need to determine the balanced chemical equation for the reaction between O2 and carbon disulfide (CS2):
2CS2 + 5O2 → 2CO2 + 4SO2
The coefficients in the balanced equation tell us the ratio in which the reactants and products combine.
Now, let's calculate the volume of SO2 produced using the given information:
1. Convert the volume of O2 given (27.9 mL) to moles by using the ideal gas law (PV = nRT). The gas constant (R) is 0.0821 L ⋅ atm/mol ⋅ K, and the temperature (T) is usually assumed to be room temperature, which is 298 K. The pressure (P) is not given, so we can assume it is at standard atmospheric pressure, 1 atm.
Using the equation PV = nRT, we can rearrange it to solve for n (the number of moles):
n = (PV) / (RT)
n = (1 atm * 0.0279 L) / (0.0821 L ⋅ atm/mol ⋅ K * 298 K)
n ≈ 0.00119 mol
2. Now, using the balanced chemical equation, we can determine the moles of SO2 produced. According to the equation, 1 mol of O2 reacts to produce 4 mol of SO2.
Therefore, using the molar ratio, the number of moles of SO2 produced will be:
0.00119 mol O2 * (4 mol SO2 / 1 mol O2) = 0.00476 mol SO2
3. Finally, convert the moles of SO2 to volume using the ideal gas law:
V = nRT / P
V = (0.00476 mol * 0.0821 L ⋅ atm/mol ⋅ K * 298 K) / 1 atm
V ≈ 0.119 L
To convert liters to milliliters, multiply by 1000:
Volume of SO2 = 0.119 L * 1000 mL/L
Volume of SO2 ≈ 119 mL
Therefore, approximately 119 milliliters of sulfur dioxide will be produced when 27.9 mL of O2 reacts with carbon disulfide (CS2).
To determine the volume of sulfur dioxide (SO2) produced when 27.9 mL of O2 reacts with carbon disulfide (CS2), we first need to balance the chemical equation:
CS2 + 3O2 --> CO2 + 2SO2
The balanced equation shows that 3 moles of O2 react to produce 2 moles of SO2. Now we can use this ratio to find the volume of SO2 produced.
1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 L (22,400 mL). Therefore, to convert moles to milliliters, we can use the ratio:
2 moles SO2 = 2 * 22,400 mL = 44,800 mL
Now we can set up a proportion:
(3 moles O2 / 27.9 mL O2) = (2 moles SO2 / x mL SO2)
Cross-multiplying, we get:
(3 moles O2) * (x mL SO2) = (27.9 mL O2) * (2 moles SO2)
3x = 55.8
Solving for x, we find:
x = 55.8 / 3
x ≈ 18.6 mL SO2
Therefore, approximately 18.6 mL of sulfur dioxide will be produced when 27.9 mL of O2 reacts with carbon disulfide.