What volume of oxygen, measured at room temperature and pressure, is required for the complete combustion of 8.65 g of H2S? Give your answer to two decimal places.

Question

abb0t · Answer

The combustion of H$_2$S is seen by the chemical equation: $\sf \color{red}{H_2S + O_2 \rightarrow H_2O + SO_2}$ 
it is somewhat similar to the combustion of an organic species. 
Now, can you balance that equation for me, please to continue.

anonymous · Answer

$$2H _{2}S +3O _{2} \rightarrow 2SO _{2} + 2H _{2}O$$

abb0t · Answer

Good, now are you familiar with dimensional analysis?
This, in general, make a plan:: $\sf \color{blue}{given \rightarrow want}$
Next, plan the route you will take: $\sf \color{green}{8.65~g~H_2S \rightarrow moles~H_2S \rightarrow moles~O_2 \rightarrow volume~O_2}$

remember, to get volume, L, use STP, which is: $\sf \large\color{red}{22.4 \frac{L}{mol}}$

abb0t · Answer

First step, is: $\sf \color{green}{8.65g~H_2S \times \frac{1~mol~H_2S}{34.08~g~H_2S}}$ 
34.08 $\sf \frac{g}{mol}$  is the MOLAR MASS of sulfane (H$_2$S)
You are multiplying, because, as you remember from algebra, the two units cancel, and you're left with moles! Which you can use to convert and cancel again to get units you want! This is where you can apply your balanced chemical equation.

Keep in mind that it's a 2:3 ratio, so you will next divide by 2 and multiply by 3, to get moles of O$_2$

anonymous · Answer

thank u