Stoichiometry Practice Hydrogen sulfide gas reacts with oxygen gas to form sulfur dioxide and water according to the following equation. 2H_(2)S(g)+3O_(2)(g)arrow 2SO_(2)(g)+2H_(2)O(g) What volume of hydrogen sulfide is needed to react completely with 4.2 L oxygen? 4.2 L 8.4 L 1.9 L 2.8L

## Step 1<br />The balanced chemical equation given is:<br />### \(2H_{2}S(g) + 3O_{2}(g) \rightarrow 2SO_{2}(g) + 2H_{2}O(g)\)<br />This equation tells us that 2 moles of hydrogen sulfide (H2S) react with 3 moles of oxygen (O2) to produce 2 moles of sulfur dioxide (SO2) and 2 moles of water (H2O).<br /><br />## Step 2<br />The stoichiometry of the reaction is 2:3, which means that for every 3 moles of oxygen, 2 moles of hydrogen sulfide are required.<br /><br />## Step 3<br />The problem provides that 4.2 L of oxygen is available. Since the stoichiometry of the reaction is 2:3, we can set up a proportion to find the volume of hydrogen sulfide needed.<br /><br />## Step 4<br />The proportion is set up as follows:<br />### \(\frac{2 \text{ moles of } H_{2}S}{3 \text{ moles of } O_{2}} = \frac{x \text{ moles of } H_{2}S}{4.2 \text{ L of } O_{2}} \)<br /><br />## Step 5<br />Solving for x gives us the volume of hydrogen sulfide needed.