Question

16 H2S(g) + 8 SO2(g) → 3 S8(s) + 16 H2O(g)

What volume of hydrogen sulfide gas, H2S, at 24.4 °C and 759 mmHg will produce 5.85 g of Sulfur? The molar mass of S8 is 256.56 g/mol.

Answer 1

Therefore, about 2.79 L of H2S gas at 24.4 °C and 759 mmHg is needed to produce 5.85 g of Sulfur.

Step-by-step explanation:

We can use stoichiometry to relate the amount of H2S needed to produce 5.85 g of S8:

16 moles of H2S + 8 moles of SO2 → 3 moles of S8 + 16 moles of H2O

From the balanced equation, we can see that 3 moles of S8 are produced for every 16 moles of H2S used. The molar mass of S8 is 256.56 g/mol, so 5.85 g of S8 is equivalent to:

5.85 g S8 × (1 mol S8/256.56 g S8) = 0.0228 mol S8

Since the ratio of H2S to S8 is 16:3, we can find the amount of H2S needed using the following proportion:

16 moles H2S / 3 moles S8 = x moles H2S / 0.0228 moles S8

Solving for x, we get:

x = (16/3) × 0.0228 moles H2S = 0.1224 moles H2S

Now we can use the ideal gas law to find the volume of H2S needed:

PV = nRT

V = nRT/P

where P is the pressure in atm, n is the number of moles, R is the gas constant (0.08206 L·atm/(mol·K)), and T is the temperature in Kelvin. Converting the temperature to Kelvin, we have:

T = 24.4 °C + 273.15 = 297.55 K

We are given the pressure as 759 mmHg, which is equivalent to:

759 mmHg × (1 atm/760 mmHg) = 0.998 atm

Substituting in the values, we get:

V = (0.1224 mol) × (0.08206 L·atm/(mol·K)) × (297.55 K) / (0.998 atm)

V ≈ 2.79 L