Answer:
Option B. 4 moles of the gaseous product
Explanation:
Data obtained from the question include:
Initial volume (V1) = V
Initial number of mole (n1) = 2 moles
Final volume (V2) = 2V
Final number of mole (n2) =..?
Applying the Avogadro's law equation, we can obtain the number of mole of the gaseous product as follow:
V1/n1 = V2/n2
V/2 = 2V/n2
Cross multiply
V x n2 = 2 x 2V
Divide both side by V
n2 = (2 x 2V)/V
n2 = 2 x 2
n2 = 4 moles
Therefore, 4 moles of the gaseous product were produced.
Convert the mass to moles .
85.1 g ÷ 20.18 g/mol = 4.21704658
convert the moles to molecules
4.2170 mol × 6.022^23 molecules/mol = 2.539^24
Answer : The mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.
Explanation : Given,
Mass of oxygen in sulfur dioxide = 3.49 g
Mass of sulfur in sulfur dioxide = 3.50 g
Mass of oxygen in sulfur trioxide = 9.00 g
Mass of sulfur in sulfur trioxide = 6.00 g
Now we have to calculate the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide.
Mass of oxygen per gram of sulfur for sulfur dioxide = 
Mass of oxygen per gram of sulfur for sulfur dioxide = 
and,
Mass of oxygen per gram of sulfur for sulfur trioxide = 
Mass of oxygen per gram of sulfur for sulfur trioxide = 
Thus, the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.
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