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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

Respuesta :

axrose
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:

The correct answer is option B.

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

[tex]PV=nRT[/tex]

[tex]n=\frac{0.99 atm\times 4.5 L}{0.0821 atm L/mol K\times 573.15 K}[/tex]

n = 0.09533 mol

[tex]S+O_2\rightarrow  SO_2[/tex]

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

Then 0.09533 mol will be obtained from:

[tex]\frac{1}{1}\times 0.09533 mol=0.09533 mol[/tex]

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

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