# Question #e4d8f

##### 1 Answer

You'll need about

It depends on what you mean by

What we'll first need to do is get the mols of

#("6 g H"_2"SO"_3)/"100 g soln" ~~ ("6 g H"_2"SO"_3)/"100 g water"#

At

#(6 cancel("g H"_2"SO"_3))/(100 cancel"g water") xx (992.2187 cancel"g soln")/("1 L soln") xx ("1 mol H"_2"SO"_3)/(82.07 cancel"g")#

#~~# #"0.7254 M"#

In order to release

#n = (PV)/(RT) = (("1 atm")("0.2 dm"^3))/(("0.082057 dm"^3cdot"atm/mol"cdot"K")("313.15 K")) ~~ "0.00778 mols SO"_2#

But the most *accurate* way is to get the density of

#0.2 cancel("dm"^3) xx cancel"1 L"/cancel("1 dm"^3) xx (2.5031 cancel("g SO"_2))/cancel"L" xx "1 mol SO"_2/(64.066 cancel("g SO"_2))#

#~~ "0.00781 mols SO"_2#

Not too far off, but I'm going with the density! (Besides, if you go off the assumption that you need more mols of

From the stoichiometry of the reaction, that requires

#("L")/(0.7254 cancel("mols H"_2"SO"_3)) xx 0.00781 cancel("mols H"_2"SO"_3) ~~ "0.01078 L"#

Or about