# Question e4d8f

Jul 27, 2017

You'll need about $\text{10.78 mL}$ of ${\text{H"_2"SO}}_{3}$, if you have 6% "w/w" ${\text{H"_2"SO}}_{3}$.

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 ${\text{H"_2"SO}}_{3}$ 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 ${40}^{\circ} \text{C}$, the density of water is about $\text{0.9922187 g/mL}$, 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")

$\approx$ $\text{0.7254 M}$

In order to release ${\text{0.2 dm}}^{3}$ of ${\text{SO}}_{2} \left(g\right)$ 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 ${\text{SO}}_{2} \left(g\right)$ at ${40}^{\circ} \text{C}$, which is about $\text{2.5031 g/L}$ (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))

$\approx {\text{0.00781 mols SO}}_{2}$

Not too far off, but I'm going with the density! (Besides, if you go off the assumption that you need more mols of ${\text{H"_2"SO}}_{3}$ than you actually do, you won't run out of ${\text{H"_2"SO}}_{3}$ before you get ${\text{0.2 dm}}^{3}$ of ${\text{SO}}_{2}$.)

From the stoichiometry of the reaction, that requires $\text{0.00781 mols}$ of ${\text{H"_2"SO}}_{3}$. So, you'll need this volume of ${\text{H"_2"SO}}_{3}$:

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

Or about $\textcolor{b l u e}{\text{10.78 mL}}$.