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To produce 40.0 g of silver chromate, you will need at least 23.4 g of potassium chromate in solution as a reactant. All you have on hand in the stock room is 5 L of a 6.00 M K2CrO4 solution. What volume of the solution is needed to give you the 23.4 g K2CrO4 needed for the reaction?

Respuesta :

Answer:

volume of [tex]K_2CrO_4[/tex] required=20.1 ml

Explanation:

First calculate the number of mole of [tex]K_2CrO_4[/tex],

given mass of [tex]K_2CrO_4[/tex]=23.4 gram ⇒this is the required mass to produce 40 gram silver chromate

molecular weight=194g/mol

[tex]mole=\frac{given \, mass}{molecular\, weight}[/tex]

mole=0.121 mol

molarity is given and we have also calculated the mole so we will use the relation between molarity,volume and mole i.e.

[tex]molarity= \frac{mole}{volume\, in\, L}[/tex]

[tex]6= \frac{0.121}{volume\, in\, L}[/tex]

volume in L=0.0201 lire

volume of [tex]K_2CrO_4[/tex] required=20.1 ml

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