Question

1. What is the maximum mass of iron that can be obtained by the reaction of 65.25 g Fe₂O₃ and 25.96 g of H₂?

Fe₂O_3(s) + 3H_2(g) -> 2 Fe_(s) + 3H₂O_(g)

2. What is the percentage yield for a reaction in which 2.50g of AgCl is obtained from 1.50g of strontium chlorida and a slight excess of silver nitrate?

SrCl_2(aq) + 2AgNO_3(aq) -> 2AgCl_(s) + Sr(NO₃)_{2 (aq)}

3. Determine the oxidation numbers for each chemical species. identify the reduced and oxidized chemical species.

2PbO_(s) + PbS_(s) -> 3Pb_(s) + SO_2(g)

Answer 1

1: Given mass of Fe2O3 = 65.25 g

Molecular mass of Fe2O3 = 159.69 g/mol

Hence moles of Fe2O3 = mass / molecular mass = 65.25 g / 159.69g/mol = 0.4086 mol

Moles of H2(g) = 25.96 g / 2 g/mol = 12.98 mol

The balanced reaction is

Fe2O3(s) + 3H2(g) -> 2 Fe(s) + 3H2O(g)

In the above balanced reaction it is clear that 1 mole of Fe2O3 reacts with 3 mol of H2.

Hence 0.4086 mol of Fe2O3 that will react with the moles of H2 = 0.4086 mol Fe2O3 x (3 mol H2 / 1 mol Fe2O3)

= 1.226 mol H2

Hence Fe2O3 is exhausted completely reacting with only 1.226 mol of H2. Hence Fe2O3 is the limiting reactant and decides the amount of products formed.

Also in the above balanced reaction, 1 mole of Fe2O3 forms 2 moles of Fe

Hence  0.4086 mol of Fe2O3 that will form the moles of Fe = 0.4086 molFe2O3 x (2 mol Fe / 1 mol Fe2O3)

= 0.8172 mol Fe

Hence mass of Fe(s) formed = 0.8172 mol Fe x 55.84 g/mol = 45.63 g (answer)

Q:2: Moles of SrCl2 = mass / molecuar mass = 1.50 g / 158.53 g/mol = 0.009462 mol

The balanced reaction is

SrCl2(aq) + 2AgNO3(aq) -> 2AgCl(s) + Sr(NO3)2 (aq)

In the above balanced reaction, 1 mol of SrCl2 forms 2 mol of AgCl

Hence 0.009462 mol of SrCl2 that will form the moles of AgCl

= 0.009462 mol SrCl2 x (2 mol AgCl / 1 mol SrCl2) = 0.018924 mol AgCl

Molecular mass of AgCl = 143.32 g/mol

Hence mass of AgCl formed =  0.018924 mol x 143.32 g/mol = 2.71 g

Hence theoritical mass of AgCl = 2.71 g

Actual mass of AgCl formed = 2.50 g

Hence percentage yield = (2.50 g/ 2.71g) x 100 = 92.2 % (answer)

3: 2PbO(s) + PbS(s) -> 3Pb(s) + SO2(g)

ON of Pb in PbO = +2

ON of O in PbO and SO2 = - 2

ON of Pb in PbS = +2

ON of S in PbS = - 2

ON of Pb(s) = 0

ON of S in SO2 = +4

Here Pb in both PbO and PbS is reduced from +2 to 0

S in PbS is oxidised to SO4 from -2 to +4