# Hydrogen sulfide reacts with sulfur dioxide to give H2O and S, H2S + SO2 = H2O + S(solid), unbalanced. If 3.0 L of H2S gas at 760 torr produced 4.8 g of sulfur, what is the the temperature in Celsius?

Jul 13, 2014

The temperature of the gas is 90 °C.

#### Explanation:

You must convert

$\text{grams of S" → "moles of S" → "moles of H"_2"S}$

and then use the Ideal Gas Law to find the temperature of the $\text{H"_2"S}$.

1. Write the balanced equation.

The balanced equation is

$\text{2H"_2"S" + "SO"_2 → "2H"_2"O" + "3S}$

2. Calculate the moles of $\text{H"_2"S}$.

$\text{moles of H"_2"S" = 4.8 color(red)(cancel(color(black)("g S"))) × (1color(red)(cancel(color(black)("mol S"))))/(32.06color(red)(cancel(color(black)("g S")))) × ("2 mol H"_2"S")/(3color(red)(cancel(color(black)("mol S")))) = "0.0998 mol H"_2"S}$

3. Use the Ideal Gas Law to calculate the temperature of the $\text{H"_2"S}$.

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

P = 760 color(red)(cancel(color(black)("Torr"))) × "1 atm"/(760color(red)(cancel(color(black)("Torr")))) =" 1.00 atm"

T = (PV)/(nR) = (1.00 color(red)(cancel(color(black)("atm"))) × 3.0color(red)(cancel(color(black)("L"))))/(0.0998color(red)(cancel(color(black)("mol"))) × "0.082 06"color(red)(cancel(color(black)("L·atm")))·"K"^"-1"color(red)(cancel(color(black)("mol"^"-1")))) = "366 K"

$T = \text{(366 – 273.15) K" = "90 °C}$

Note: The answer can have only 1 significant figure, because the Kelvin temperature has only 2 significant digits. If you need more precision, you will have to recalculate.