In the laboratory, a student adds 19.7 g of barium acetate to a 500. ml volumetric flask and adds water to the mark on the neck of the flask. calculate the concentration (in mol/l) of barium acetate, the barium ion and the acetate ion in the solution. [ba(ch3coo)2] = m [ba2+] = m [ch3coo-] = m
calculate the mass, in grams, of iron(ii) sulfate that must be added to a 300-ml volumetric flask in order to prepare 300 ml of a 0.181 m aqueous solution of the salt.
grams

1)

2) 21.72 grams of  iron(II) sulfate that must be added.

Explanation:

1) Moles of barium acetate =

Volume of the solution was made to 500 ml that 0.5 L

In 1 mole of barium acetate there are 1 mole of barium ions and 2 moles of acetate ions.

2) Moles of iron(II) sulfate be n

Volume of the solution = 300 mL= 0.3 L

n = 0.0543 moles

Mass of 0.0543 moles of iron(II) sulfate:

0.0543 mol × 400 g/mol = 21.72 g

21.72 grams of  iron(II) sulfate that must be added.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"