# If 96 g SO2 is added to 2 moles of oxygen at STP calculate the total volume of gas in the container at the end of the reaction

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1. 2SO2 + O2 ==> 2SO3
This is a limiting reagent (LR) problem. The non LR is called the ER for excess reagent.
mols SO2 = grams/molar mass = 96/64 = 1.5
How much O2 will be required to use all of the SO2. That's
1.5 mols SO2 x (1 mol O2/2 mols SO2) = 1.5 x 1/2 = 0.75. Do you have that much O2. Yes so SO2 is the LR and O2 is the ER.
How much SO3 is formed. That's 1.5 mols SO2 x (2 mols SO3/2 mols SO2) = 1.5 mols SO3 formed.
How much O2 is used. That's above @ 0.75 moles.
So you have 2 moles O2 initially - 0.75 mols O2 used in the reaction to leave an excess of 1.25 mols O2 unused. Total gas then is moles SO3 formed (1.5) + mols O2 left unreacted (1.25) = total mols gas after the reaction.
Final volume = moles SO3 x 22.4 L/mol@STP = ?

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2. 2SO2(g)+O2(g)>2SO3(g)
RFM of SO2= 32+2×16=64
no of mole of SO2 =96/64=1.5moles
mole ratio of SO2 to SO3 is 2:2, thus no of moles of SO3 is 1.5moles
1 mole at STP=22.4Litres
1.5 moles=??
=1.5×22.4
=33.6 Litres

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