Question

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?

Answer 1

Answer:

volume of $K_2CrO_4$ required=20.1 ml

Explanation:

First calculate the number of mole of $K_2CrO_4$,

given mass of $K_2CrO_4$=23.4 gram ⇒this is the required mass to produce 40 gram silver chromate

molecular weight=194g/mol

$mole=\frac{given \, mass}{molecular\, weight}$

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.

$molarity= \frac{mole}{volume\, in\, L}$

$6= \frac{0.121}{volume\, in\, L}$

volume in L=0.0201 lire

volume of $K_2CrO_4$ required=20.1 ml