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

To obtain 23.4 g of K2CrO4, you will need 0.0201 L (or 20.1 mL) of the 6.00 M K2CrO4 solution.

Step-by-step explanation:

To find the volume of the K2CrO4 solution needed to obtain 23.4 g of K2CrO4, we can use the concentration of the solution and convert it to moles of K2CrO4. Then, we can use the molar mass of K2CrO4 to convert moles to grams. Finally, we can use the given mass of K2CrO4 to find the volume of the solution needed.

First, we calculate the number of moles of K2CrO4 needed:

moles of K2CrO4 = mass / molar mass = 23.4 g / (194.19 g/mol) = 0.1203 mol

Next, we can use the molarity of the solution to find the volume of the solution needed:

volume of solution = moles / molarity = 0.1203 mol / 6.00 M = 0.0201 L

Therefore, you will need 0.0201 L (or 20.1 mL) of the 6.00 M K2CrO4 solution to obtain 23.4 g of K2CrO4.

Answer 2

volume of
`K_2CrO_4` required=20.1 ml

Step-by-step 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=(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= (mole)/(volume\, in\, L)`

`6= (0.121)/(volume\, in\, L)`

volume in L=0.0201 lire

volume of
`K_2CrO_4` required=20.1 ml