Question #d8775

1 Answer
May 28, 2015

Your reaction will produce 150. mL of carbon dioxide and 300. mL of sulfur dioxide under STP conditions.

Since you didn't provide the conditions for pressure and temperature, I"ll assume your reaction to take place at STP - Standard Temperature and Pressure.

STP conditions imply a pressure of 100 kPa and a temperature of 273.15 K. Under these conditions, 1 mole of any ideal gas occupies exactly 22.7 L - this is known as the molar volume of a gas at STP.

Take a look at the balanced chemical equation for your reaction

#CS_(2(l)) + color(red)(3)O_(2(g)) -> CO_(2(g)) + color(blue)(2)SO_(2(g))#

Since no mention of a specific amount of carbon disulfide taking part in the reaction was made, you can assume it to be in excess. This means that all the moles of oxygen will take part in the reaction.

Notice the mole ratios that exist between oxygen and the two products. For every #color(red)(3)# moles of oxygen that react, you get 1 mole of #CO_2# and #color(blue)(2)# moles of sulfur dioxide.

So, use the molar volume of a gas at STP to determine how many moles of oxygen took part in the reaction.

#450. * 10^(-3)cancel("L") * "1 mole"/(22.7cancel("L")) = "0.0198 moles"# #O_2#

This means that the reaction produced

#0.0198cancel("moles"O_2) * ("1 mole"CO_2)/(color(red)(3)cancel("moles"O_2)) = "0.00660 moles"# #CO2#


#0.0198cancel("moles"O_2) * (color(blue)(2)"moles"SO_2)/(color(red)(3)cancel("moles"O_2)) = "0.0132 moles"# #SO_2#

To determine the volume of each gas produced, use the STP molar volume again

#0.00660cancel("moles") * "22.7 L"/(1cancel("mole")) = "0.1498 L"#


#0.0132cancel("moles") * "22.7 L"/(1cancel("mole")) = "0.2996 L"#

Expressed in mL and rounded to three sig figs, the number of sig figs you gave for the volume of oxygen that reacted, the answer will be

#V_(CO_2) = color(green)("150. mL")#

#V_(SO_2) = color(green)("300. mL")#

SIDE NOTE Many chemistry textbooks and online sources still use the old definition of STP, which is a pressure of 1 atm and a temperature of 273.15 K. Under these conditions, the molar volume of a gas is equal to 22.4 L.

If you wish, you can redo the calculations using the old value.