Question

2X(s) + 6HCl (aq) ----> 2 XCl3 (aq) + 3 H2(g)

a. Which reactant is oxidized and which reactant is reduced? (Please explain)

b. To prepare a HCl solution, 12.0 M concentrated HCl was diluted with water to make 250 mL of a 4.00 M HCl solution. How much, in mL, of concetrated acid was used?

c. In an experiment, it was found that 25 mL of the diluted acid was required to react completely with the 4.80 g of X. What is element X?

Answer 1

Ans :

a)

X has 0 oxidation state in reactant side and +3 in products side , so X is getting oxidised.

H has +1 oxidation state in reactant side and 0 in products side , so H is getting reduced.

b)

We can use the formula , M1V1 = M2V2

M1=12.0 M

M2 = 4.00 M

V2 = 250 mL

Putting values :

12.0 x V1 = 4.00 x 250

V1 = 83.33 mL

So 83.33 mL of concentrated acid was used.

c)

Number of moles of acid = molarity x volume (L)

= 4.00 x 0.025

= 0.1 mol

6 mol HCl uses 2 mol of X

So 0.1 mol of HCl will use : (0.1 x 2) / 6 = 0.033 mol X

Molar mass of X = given mass / number of mol

= 4.80 / 0.033

= 144 g/mol

The molar mass corresponds to the element Neodymium . So the element X is Neodymium.