Chat with us, powered by LiveChat In this reaction, 2SO2(g)+O2(g)→2SO3(g), 284.9mL of SO2 is allowed to react with 159.0mL of O2 (measured at 330K and 48.5mmHg ). What is theoretical yield of SO3? If 174.3mL of SO3 is collected (measured at 330K and 48.5mmHg ), what is the percent yield? - Wridemy

## 14 Aug In this reaction, 2SO2(g)+O2(g)→2SO3(g), 284.9mL of SO2 is allowed to react with 159.0mL of O2 (measured at 330K and 48.5mmHg ). What is theoretical yield of SO3? If 174.3mL of SO3 is collected (measured at 330K and 48.5mmHg ), what is the percent yield?

The theoretical yield of sulfur trioxide, or ##SO_3##, will be 285.0 mL, and the will be 61.16%.

##color(red)(2)SO_(2(g)) + O_(2(g)) -> color(red)(2)SO_(3(g))##

Notice that you have a ##color(red)(2):1## between sulfur dioxide, ##SO_2##, and oxygen; this will be very important when comparing the number of moles of each gas that react, since it will tell you if one of the two will act as a .

Your tool of choice will be the equation. Start by determining the number of moles of sulfur dioxide and oxygen that you start with

##PV = nrt => n = (PV)/(RT)##

##n_(O_2) = (48.5/760cancel(“atm”) * 159 * 10^(-3)cancel(“L”))/(0.082 (cancel(“L”) * cancel(“atm”))/(“mol” * cancel(“K”)) * 330cancel(“K”)) = “0.0003750 moles “## ##O_2##

##n_(SO_2) = (48.5/760cancel(“atm”) * 284.9 * 10^(-3)cancel(“L”))/(0.082 (cancel(“L”) * cancel(“atm”))/(“mol” * cancel(“K”)) * 330cancel(“K”)) = “0.0006719 moles “## ##SO_2##

According to ratio, that many moles of oxygen would have required

##0.0003750cancel(“moles “O_2) * (color(red)(2)”moles “SO_2)/(1cancel(“mole “SO_2)) = “0.000750 moles “## ##SO_2##

Since you’ve got fewer moles than required, sulfur dioxide will act as a , i.e. it will determine how many moles of oxygen will actually react

##0.0006719cancel(“moles “SO_2) * (“1 mole “O_2)/(color(red)(2)cancel(“moles “SO_2)) = “0.0003360 moles “O_2##

The number of moles of sulfur trioxide this reaction will produce is

##0.0006719cancel(“moles “SO_2) * (“1 mole “SO_3)/(1cancel(“mole “SO_2)) = “0.0006719 moles “SO_3##

This many moles will produce a volume of

##V = (nRT)/P = (0.0006719cancel(“moles”) * 0.082((“L” * cancel(“atm”))/(cancel(“mol”) * cancel(“K”)) * 330cancel(“K”)))/(48.5/760cancel(“atm”))##

##V = “0.285 L” = color(green)(“285.0 mL”)##

This represents the theoretical yield of ##SO_3## ##->## this is how much sulfur trioxide your reaction would produce if all (100%) of the sulfur dioxide reacts.

SInce your reaction produced 174.3 mL, the of the reaction will be smaller than 100%. Calculate by

##”% yield” = “actual yield”/”theoretical yield” * 100##

##”% yield” = 174.3/285.0 * 100 = color(green)(“61.16%”)##

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