# Question #85af7

Feb 24, 2015

First things first, your equation is mistyped, since you have oxygen on the reactans' side, but not on the products' side. The correct balanced equation for this reaction is this one

$8 {H}_{2 \left(g\right)} + {S}_{8 \left(g\right)} \to 8 {H}_{2} {S}_{\left(g\right)}$

Now for the mole ratio between ${H}_{2}$ and ${H}_{2} S$. What a mole ratio tells is in what proportion the two compounds will be when this reaction takes place. Notice that the coefficients in front of ${H}_{2}$ and ${H}_{2} S$ are both equal to 8.

This is what a $\text{1:1}$ mole ratio is - the number of moles of one compounds is identical to the number of moles of another compound.

What that means is that regardless of how many moles of ${H}_{2}$ you have, the number of moles of ${H}_{2} S$ produced will always be equal to the number of moles of ${H}_{2}$ that react.

If you have, for example, 0.100 moles of ${H}_{2}$ that react, you'll automatically have 0.100 moles of ${H}_{2} S$ that are produced.

By comparison, the mole ratio between ${H}_{2}$ and ${S}_{8}$ is $\text{8:1}$. Regardless of how many moles of ${S}_{8}$ react, you'll always have 8 times more moles of ${H}_{2}$ that are needed to react.

In essence, a mole ratio tells you the proportion in which the species involved in a reaction must be at all times.

So, if I give you 0.0250 moles of ${H}_{2}$ and ask you to tell me the number of moles of ${S}_{8}$ you need to react with this amount of hydrogen and the number of moles of ${H}_{2} S$ you produce, use the mole ratios.

$\text{0.0250 moles"H_2 * ("1 mole"S_8)/("8 moles"H_2) = "0.00313 moles}$ ${S}_{8}$

$\text{0.0250 moles"H_2 * ("1 mole"H_2)/("1 mole"H_2S) = "0.0250 moles}$ ${H}_{2} S$