3.8 liters of sulfur vapor, S8(g), at 921.4°C and 5.87 atm is burned in excess pure oxygen gas to give sulfur dioxide gas measured at the same temperature and pressure. What mass (in g) of sulfur dioxide gas is obtained? Enter to 1 decimal place.

116.5 g of SO₂ are formed

Explanation:

The reaction is:

S₈(g) +  8O₂(g)  → 8SO₂ (g)

Let's identify the moles of sulfur vapor, by the Ideal Gases Law

We convert the 921.4°C to Absolute T° → 921.4°C + 273 = 1194.4 K

5.87 atm . 3.8L = n . 0.082 L.atm/mol.K . 1194.4K

(5.87 atm . 3.8L) / (0.082 L.atm/mol.K . 1194.4K) = n → 0.228 moles of S₈

Ratio is 1:8, 1 mol of sulfur vapor can produce 8 moles of dioxide

Then, 0.228 moles of S₈ must produce (0.228 . 8) /1 =  1.82 moles

We convert the moles to g → 1.82 moles . 64.06 g /1mol = 116.5 g

We obtain 116.6 grams of SO2

Explanation:

Step 1: Data given

Volume of S8 = 3.8 L

Temperature = 921.4 °C = 1194.55 K

Pressure = 5.87 atm

Step 2: The balanced equation

S8 + 8O2 → 8SO2

Step 3: Calculate moles S8

pV = nRT

⇒ with p = the pressure of S8 = 5.87 atm

⇒ with V = the volume of S8 = 3.8L

⇒ with n = the number of moles S8 = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 1194.55 K

n = (pV)/(RT)

n = (5.87*3.8)/(0.08206*1194.55 )

n = 0.2276 moles

Step 4: Calculate moles SO2

For 1 mol S8 we need 8 moles O2 to produce 8 moles SO2

For 0.2276 moles S8 we'll have 8*0.2276 = 1.82 moles SO2

Step 5: Calculate mass SO2

Mass SO2 = moles SO2 * molar mass SO2

Mass SO2 = 1.82 moles *64.07 g/mol

Mass SO2 = 116.6 grams

We obtain 116.6 grams of SO2

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm