Question #e4d8f

You'll need about #"10.78 mL"# of #"H"_2"SO"_3#, if you have #6% "w/w"# #"H"_2"SO"_3#.

It depends on what you mean by #5-6%#, but I assume you mean #%"w/w"#. I will also assume #6%#, but you will have to know what the exact number is if you want a more accurate result.

What we'll first need to do is get the mols of #"H"_2"SO"_3# that would decompose. Since #6%"w/w"# is a fairly small concentration,

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

At #40^@ "C"#, the density of water is about #"0.9922187 g/mL"#, so assuming the solution density is that of water, you have a molarity of about:

#(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 #"0.2 dm"^3# of #"SO"_2(g)# in the reaction, you'll need to form however many mols that corresponds to. You could do that from the ideal gas law:

#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 #"SO"_2(g)# at #40^@ "C"#, which is about #"2.5031 g/L"# (extrapolated from here).

#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 #"H"_2"SO"_3# than you actually do, you won't run out of #"H"_2"SO"_3# before you get #"0.2 dm"^3# of #"SO"_2#.)

From the stoichiometry of the reaction, that requires #"0.00781 mols"# of #"H"_2"SO"_3#. So, you'll need this volume of #"H"_2"SO"_3#:

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

Or about #color(blue)("10.78 mL")#.