Answer:
112mL
Explanation:
We'll begin by calculating the number of mole in 0.640g of SO2.
This is illustrated below:
Molar mass of SO2 = 32 + (16x2) = 64g/mol
Mass of SO2 = 0.640g
Number of mole of SO2 =.?
Mole = mass /molar mass
Number of mole of SO2 = 0.640/64
Number of mole of SO2 = 0.01 mole
Next, we shall determine the number of mole of O2 required for the reaction. This is illustrated below:
2SO2(g) + O2(g) —> 2SO3(g)
From the balanced equation above,
2 moles of SO2 reacted with 1 mole of O2.
Therefore, 0.01 mole of SO2 will react with = (0.01 x 1)/2 = 0.005 mole of O2.
Therefore, 0.005 mole of O2 is required for the reaction.
Finally, we shall determine the volume of O2 required for the reaction as follow:
Note: 1 mole of a gas occupy 22.4L (22400mL) at stp.
1 mole of O2 occupy 22400mL at stp.
Therefore, 0.005 mole of O2 will occupy = 0.005 x 22400 = 112mL
Therefore, 112mL of O2 is required for the reaction.
