Question

Hydrogen sulfide reacts with sulfur dioxide to give \[{H_2}O\] and \[S\], \[{H_2}S + S{O_2} = {H_2}O + S\left( {solid} \right)\] , unbalanced. If \[3.0L\] of \[{H_2}S\] gas at \[760torr\] produced \[4.8g\] of sulphur, what is the temperature in Celsius?

Answer 1

Hint: The balanced chemical equation gives the number of moles of each reactant and product. Given pressure and volume. The universal gas constant value is constant. From the ideal gas law, the value of temperature can be converted into Celsius as the temperature obtained will be in kelvins.

Complete answer:
Hydrogen sulphide reacts with sulphur dioxide to give water and sulphur. The balanced chemical equation will be written as:
\[2{H_2}S + S{O_2} \to 2{H_2}O + 3S\]
Given that \[3.0L\] of \[{H_2}S\] gas at \[760torr\]produced \[4.8g\] of sulphur
Moles of \[{H_2}S\] will be \[4.8gS \times \dfrac{{1moleS}}{{32.06gS}} \times \dfrac{{2moles{H_2}S}}{{3moleS}}\] which is equal to \[0.0998\] moles
Thus, moles of \[{H_2}S\] were \[0.0998\] moles
The ideal gas constant R has the value of \[0.0826Latm{\left( {K.mol} \right)^{ - 1}}\]
From the ideal gas law, \[PV = nRT\]
The temperature will be \[T = \dfrac{{PV}}{{nR}}\]
Given pressure is in torrents, convert into atm
\[P = 760torr \times \dfrac{1}{{760}} = 1atm\]
Substitute all the values of pressure, volume, number of moles and ideal gas constant in the ideal gas equation
\[T = \dfrac{{1atm \times 3L}}{{0.0998mol \times 0.0826Latm{{\left( {K.mol} \right)}^{ - 1}}}} = 364K\]
The temperature obtained is in kelvins as kelvin is the standard unit of temperature
Convert this temperature into Celsius \[364 - 273 = {91^0}C\] as one degree centigrade is equal to \[273K\]
Thus, the temperature in Celsius is \[{91^0}C\].

Note:
The moles of one compound in a chemical reaction can be determined from the moles of other reactants in a chemical reaction. The molar mass and moles were needed to calculate the moles. The units of all terms in the ideal gas equation must be clear. The units and ideal gas constant value should be \[0.0826Latm{\left( {K.mol} \right)^{ - 1}}\]