Consider this reaction. What volume of oxygen gas, in milliliters, is required to react with 0.640 g of SO2 gas at S TP? 11.2 mL 22.4 mL 112 mL 224 mL

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.

6.4 cm (max) - 25 cm (min)