Question

10.1 Practice basic skills (grams, moles, g/mol, mol/L, dilution, stoichiometry): (a) Mow many grams of Br2 are produced when 1.50 g of FeBr; react with excess CI? 3 Cl2 + 2 FeBr + 3 Br2 + 2 FeCl3 (b) If the reaction in (a) happens in 500 mL of solution, what is the resulting molarity of CI? (c) Calculate the mL of concentrated H2SO4 (18 M) required to prepare 750 mL of 2.0 M solution.

Answer 1

a) according to equation , 3mole Br2 is produced when 2 mole of FeBr3 react with 3 mole Cl2

Given mass of FeBr3 = 1.5g

Mole of FeBr3 = given mass/molar mass

=1.5/295.56

= 0.005mole

Now since cl2 is given in excess we don't have to worry about mole of Cl2

Since 2mole FeBr3 react to produce 3 mole Br2

Then 0.005mole FeBr3 will produce x mole Br2

x = 0.005*3/2 mole

=0.0075mole Br2

One mole Br2 = 159.8g

0.0075 mole Br2 = 159.8*0.0075

=1.198g of Br2 is produced .

b) Molarity of cl- = number of mole /volume in L

Now,

According to equation 2 mole of FeBr3 produce 2 mole of FeCl3

That means equal moles

So 0.005 mole of FeCl3 is produced

One unit of Fecl3 has 3 Cl- ions

So 0.005 mole of Fecl3 will have 0.005×3 mole ions of Cl-

Volume in L = 0.5L

So molarity = 0.005×3÷0.5

= 0.03 mole/L

(c). Molarity = mole/Volume in L

Hence , no. Of mole = Molarity × volume in L

In both solutions , number of moles will remain same

Hence , M1 × V1 = M2 ×V2

18 × V1(ml) = 2 × 750 ml

V1 = 83.33 Ml