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.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2