Potassium carbonate dissolves as follows:

K2CO3(s) → 2k (aq) CO3^2-(aq).

What volume in liters of 1.0 m k2co3 solution is required to provide 0.50 moles of k (aq)?

Respuesta :

The 0.25 volume in liters of 1.0 M [tex]K_{2}CO_{3}[/tex] solution is required to provide 0.5 moles of [tex]K[/tex](aq).

Calculation,

The Potassium carbonate dissolves as follows:

[tex]K_{2}CO_{3}(s)[/tex] → [tex]2K(aq) +[/tex][tex]CO_{3}^{-2} (aq)[/tex]

The mole ratio is 1: 2

It means, the 1 mole [tex]K_{2}CO_{3}[/tex] required to form 2 mole of [tex]K[/tex](aq).

To provide 0.5 mole of [tex]K[/tex](aq) = 1 mole ×0.5 mole /2 mole required by [tex]K_{2}CO_{3}[/tex].

To provide 0.5 mole of [tex]K[/tex](aq) ,0.25 mole required by [tex]K_{2}CO_{3}[/tex].

The morality of  [tex]K_{2}CO_{3}[/tex] = 1 M = number of moles / volume in lit

The morality of  [tex]K_{2}CO_{3}[/tex] =   1 M = 0.25 mole/ volume in lit

Volume in lit = 0.25 mole / 1 M = 0.25 mole/mole/lit = 0.25 lit

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