David prepared the potassium phosphate solution by adding 46.3g to 250 mL of water. Kim needs 0.10M potassium phosphate solution. How many stock solutions does she need if she wants to mix 150 ml of diluent?

17.2mL

Explanation:

Step 1:

Determination of the molarity of potassium phosphate K3PO4 solution.

This is illustrated below:

Mass of K3PO4 = 46.3g

Molar Mass of K3PO4 = (39x3) + 31 + (16x4) = 212g/mol

Mole of K3PO4 =..?

Mole = Mass /Molar Mass

Mole of K3PO4 = 46.3/212 = 0.218 mole

Volume of water = 250mL = 250/1000 = 0.25L

Molarity = mole /Volume

Molarity of K3PO4 = 0.218/0.25

Molarity of K3PO4 = 0.872M

Step 2:

Determination of the volume of the stock solution of K3PO4 needed.

Molarity of stock solution (M1) = 0.872M

Volume of stock solution needed (V1) =..?

Molarity of diluted solution (M2) = 0.1M

Volume of diluted solution (V2) = 150mL

The volume of the stock solution needed can be obtained as follow:

M1V1 = M2V2

0.872 x V1 = 0.1 x 150

Divide both side by 0.872

V1 = (0.1 x 150)/0.872

V1 = 17.2mL

Therefore, the volume of the stock solution needed is 17.2mL

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !