Question

For the following unbalanced reaction at 1.00 atm and 27.5°C: MoS2(s) + O₂(g) → MoO₃(s) + SO₂(g), what volume of oxygen gas (in L) would be needed to react with 48.7 g of MoS2?

Answer 1

To find the volume of oxygen gas needed to react with 48.7 g of MoS2, calculate the moles of MoS2 and then use the balanced equation and ideal gas law to find the required volume, which comes out to approximately 26.2 liters.

Step-by-step explanation:

To determine the volume of O2 gas that would be needed to react with 48.7 g of MoS2, first, we need to balance the chemical equation:

2 MoS2(s) + 7 O2(g) → 2 MoO3(s) + 4 SO2(g)

Next, we need to calculate the moles of MoS2 reacting by using its molar mass:

Molar mass of MoS2 = 95.96 g/mol (Mo) + 2 * 32.07 g/mol (S) = 160.10 g/mol

Moles of MoS2 = 48.7 g / 160.10 g/mol ≈ 0.304 moles

The stoichiometry of the balanced equation shows that 2 moles of MoS2 react with 7 moles of O2. Therefore, we have:

0.304 moles MoS2 * (7 moles O2 / 2 moles MoS2) = 1.06 moles O2

At 27.5°C and 1 atm, we use the ideal gas law to find the volume of 1 mole of O2:

V = nRT / P

Where:

• n = number of moles of O2
• R = 0.0821 L atm/mol K, ideal gas constant
• T = temperature in Kelvin = 27.5°C + 273.15 = 300.65 K
• P = pressure in atm = 1 atm

Therefore, the volume needed for 1.06 moles of O2 is:

V = (1.06 moles) * (0.0821 L atm/mol K) * (300.65 K) / (1 atm) ≈ 26.2 L

So, about 26.2 liters of O2 gas is needed to react with 48.7 g of MoS2 under the given conditions.