Question

What mass of sulfur has to burn to produce 4.5L SO2 at 300°C and 101 kPa in the following reaction?

A. 13.5 g S
B. 3.07 g S
C. 68.8 g S
D. 41.0 g S

Answer 1

B. 3.07 g
You can solve this by using the formula, PV = nRT
P = Pressure
V = Volume
n = moles
R = Gas Constant
T = Temperature
You have all the variables, just solve for n. Then convert it to grams.

Answer 2

The correct answer is option B.

Step-by-step explanation:

Pressure of the gas = P = 101 kPa = 0.9968 atm

Volume of the gas = V = 4.5 L

Temperature of the gas = T = 300°C = 573.15 K

Moles of sulfur dioxide = n

`PV=nRT`

`n=(0.99 atm* 4.5 L)/(0.0821 atm L/mol K* 573.15 K)`

n = 0.09533 mol

`S+O_2\rightarrow  SO_2`

According to reaction , 1 mol sulfur dioxide is obtained from 1 mol of sulfur.

Then 0.09533 mol will be obtained from:

`(1)/(1)* 0.09533 mol=0.09533 mol`

Mass of 0.09533 moles of sulfur = 0.09533 mol × 32 g/mol= 3.07 g