Question

Question #e4d8f

Answer 1

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

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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")`.