Question

a reatus we 34. (3) What volume of Oz gas is necessary to produce 5.8 L of sulfur dioxide gas. The reaction is oxygen to produce sulfur dioxide and water. (Imol - 22.4 L at STP) 2 H2S + 302 → 2 SO2 + 2H2O a. 5.8 L b. 8.7 L c. 3.9 L d. 4.3 L

Answer 1

Let us first convert the volume of SO2 to number of moles.

As 22.4 L = 1 mol

5.8 L = (5.8 L / 22.4 L) mol

= 0.26 moles

Now from the balanced chemical equation it is clear that 3 moles of O2 produces 2 moles of SO2. Therefore, we can consider

2 moles of SO2 = 3 moles of O2

For 0.26 moles of SO2 = (0.26 x 3) / 2 moles of O2

= 0.388 moles

As, 1 mole = 22.4 L

0.388 moles = (0.388 x 22.4) L

= 8.7 L

Therefore, the answer is b. 8.7 L