SO2-> H2SO3->Li2SO3

V(SO2) = 6,82 л (н.у.)

m, v(Li2SO3) -?

To find the volume of Li2SO3, we need to use the stoichiometry of the reaction and the given volume of SO2.

From the balanced equation:
1 mole of SO2 reacts with 1 mole of H2SO3, which then reacts with 1 mole of Li2SO3.

Given that the volume of SO2 is 6.82 L (at standard conditions), we can use the ideal gas law to find the number of moles of SO2:

n(SO2) = V(SO2) / Vm(SO2)

Where Vm(SO2) is the molar volume of SO2 at standard conditions (22.4 L/mol).

n(SO2) = 6.82 L / 22.4 L/mol = 0.304 moles of SO2

Since the stoichiometry of the reaction is 1:1:1, we have 0.304 moles of H2SO3 and 0.304 moles of Li2SO3.

Now, we can find the volume of Li2SO3 using the ideal gas law:

V(Li2SO3) = n(Li2SO3) * Vm(Li2SO3)

Where Vm(Li2SO3) is the molar volume of Li2SO3 at standard conditions.

Since we don't have the molar volume of Li2SO3, we can assume it to be the same as the molar volume of SO2 (22.4 L/mol).

V(Li2SO3) = 0.304 moles * 22.4 L/mol = 6.81 L

Therefore, the volume of Li2SO3 is approximately 6.81 L.