Question

Calculations Involving a Limiting Reactant

Question 16.

This question has multiple parts. Work all the parts to get the most points.

a. "Smelling salts," which are used to revive someone who has fainted, typically contain ammonium carbonate, (NH₄)₂CO₃. Ammonium carbonate decomposes readily to form ammonia, carbon dioxide, and water. The strong odor of the ammonia usually restores consciousness in the person who has fainted. The unbalanced equation is

(NH₄)₂CO₃(s) → NH₃(g) + CO₂(g) + H₂O(g)

Calculate the mass of ammonia gas that is produced if 2.25 g of ammonium carbonate decomposes completely.

___________________ g NH3

b. Sulfurous acid is unstable in aqueous solution and gradually decomposes to water and sulfur dioxide gas (which explains the choking odor associated with sulfurous acid solutions).

H₂SO₃(aq) → H₂O(l) + SO₂(g)

If 6.50 g of sulfurous acid undergoes this reaction, what mass of sulfur dioxide is released?

_____________________ g SO2

c. For the following unbalanced chemical equation, suppose that exactly 1.75 g of each reactant is taken. Determine which reactant is limiting, and calculate what mass of CO₂ is expected (assuming that the limiting reactant is completely consumed).

CS₂(l) + O₂(g) → CO₂(g) + SO₂(g)

Limiting reactant is _______________________

__________________ g CO2

d.   Lead(II) carbonate, also called "white lead," was formerly used as a pigment in white paints. However, because of its toxicity, lead can no longer be used in paints intended for residential homes. Lead(II) carbonate is prepared industrially by reaction of aqueous lead(II) acetate with carbon dioxide gas. The unbalanced equation is

Pb(C₂H₃O₂)₂(aq) + H₂O(l) + CO₂(g) →

PbCO₃(s) + 2HC₂H₃O₂(aq)

Suppose an aqueous solution containing 1.35 g of lead(II) acetate is treated with 5.95 g of carbon dioxide. Calculate the theoretical yield of lead carbonate.

_______________________ g PbCO3

Answer 1

a. "Smelling salts," which are used to revive someone who has fainted, typically contain ammonium carbonate, (NH₄)₂CO₃. Ammonium carbonate decomposes readily to form ammonia, carbon dioxide, and water. The strong odor of the ammonia usually restores consciousness in the person who has fainted. The unbalanced equation is

(NH₄)₂CO₃(s) → NH₃(g) + CO₂(g) + H₂O(g)

The balanced equation is as follows:

heat + (NH4)2CO3--->2NH3(g)+CO2(g)+H2O

Calculate the mass of ammonia gas that is produced if 2.25 g of ammonium carbonate decomposes completely.

___________________ g NH3

Moles of ammonium carbonate in 2.25 g = 2.25 g/ 96.09 g/mol

= 0.0234 moles (NH4)2CO3

Now calculate the moles of NH3:

0.0234 moles (NH4)2CO3 * 2 moles NH3/1.00 moles (NH4)2CO3

= 0.0468 Moles NH3

Amount of NH3 = number of moles * molar mass

= 0.0468 Moles NH3 *17.031g/ mole

=0.797 g

= 0.80 g NH3

b. Sulfurous acid is unstable in aqueous solution and gradually decomposes to water and sulfur dioxide gas (which explains the choking odor associated with sulfurous acid solutions).

H₂SO₃(aq) → H₂O(l) + SO₂(g)

If 6.50 g of sulfurous acid undergoes this reaction, what mass of sulfur dioxide is released?

_____________________ g SO2

Moles of sulfurous acid in 6.50 g = 6.50 g/ 82.07 g/mol

= 0.0792 moles H₂SO₃

Now calculate the moles of SO2:

0.0792 moles H₂SO₃* 1 moles SO2/1.00 moles H₂SO₃

= 0.0792 moles SO₂

Amount of SO2 = number of moles * molar mass

= 0.0792 moles SO₂*64.066 g/ mole

=5.07 g SO2

c. For the following unbalanced chemical equation, suppose that exactly 1.75 g of each reactant is taken. Determine which reactant is limiting, and calculate what mass of CO₂ is expected (assuming that the limiting reactant is completely consumed).

CS₂(l) + O₂(g) → CO₂(g) + SO₂(g)

3 O2 + CS2 --> CO2 + 2 SO2

Moles of CS2 = 1.75 g / 76.139 g/mol

= 0.023 Moles CS2

Moles of O2 = 1.75 g / 31.99 g/mol

= 0.055 Moles O2

Limiting reactant is ______ O2_________________

Moles of CO2 =

0.055 Moles O2* 1 mole CO2/ 3 mole O2

= 0.018 Mole CO2

Amount of CO2 = 0.018 Mole CO2*44.01 g/ mole

= 0.81 g CO2

__________________ g CO2

d.   Lead(II) carbonate, also called "white lead," was formerly used as a pigment in white paints. However, because of its toxicity, lead can no longer be used in paints intended for residential homes. Lead(II) carbonate is prepared industrially by reaction of aqueous lead(II) acetate with carbon dioxide gas. The unbalanced equation is

Pb(C₂H₃O₂)₂(aq) + H₂O(l) + CO₂(g) →

PbCO₃(s) + 2HC₂H₃O₂(aq)

Pb(C₂H₃O₂)₂ + H₂O + CO₂ = PbCO₃ + 2 HC₂H₃O₂

Suppose an aqueous solution containing 1.35 g of lead(II) acetate is treated with 5.95 g of carbon dioxide. Calculate the theoretical yield of lead carbonate.

_______________________ g PbCO3

Moles of lead(II) acetate= 1.35 g/ 325.29 g/mol

= 0.00415 mole Pb(C₂H₃O₂)₂

Now calculate the moles of CO2

5.95 g of carbon dioxide /44.01 g/ mole

= 0.135 Moles CO2

Here lead(II) acetate is limiting agent now calculate the moles of PbCO3

0.00415 mole Pb(C₂H₃O₂)₂*1 mole PbCO3/1.00 mole Pb(C₂H₃O₂)₂

0.00415 mole PbCO3

Amount of PbCO3

= 0.00415 mole PbCO3 * molar mass; 267.21 g/mol

= 1.11 g PbCO3