Question

In each of the following questions, identify the limiting factor and the excess reactant, and then calculate the amount of the indicated product formed from the amounts given.

1. Given 2.0 moles of hydrogen gas and 1.6 moles of chlorine gas, calculate the moles of hydrochloric acid, HCI, produced.
H₂ + Cl2 → 2 HCI
2. Given 112.0 g of Fe and 66.7 g of water, calculate the mass of Fe304 produced. How many grams of the excess reactant remain?
3Fe(s) + 4H2O(g) —-> Fe3O4(s) +4H2(s)
3. Given 0.57 mol of octane (C8H18) and 18.1 mol of oxygen gas, calculate the number of moles of H₂O produced.
2CgH18(I) 2502(g) → 16 CO2(g)+ 18 H2O(g)

Answer 1

3.2 moles of HCl is produced

The mass of the excess reagent is 18 g

The mass of the water 92.34 g.

1) In the first reaction;

1 mole of hydrogen reacts with 1 mole of chlorine

2 moles of hydrogen reacts with 2 * 1/1

= 2 moles

Chlorine is the limiting reactant;

1 mole of chlorine produces 2 moles of HCl

1.6 moles of chlorine will produce 1.6 * 2/1

= 3.2 moles

2) Number of moles of iron = 112.0 g/56 g/mol

= 2 moles

Number of moles of water = 66.7/18 g/mol

= 3.7 moles

3 moles of iron reacts with 4 moles of water

2 moles of iron reacts with 2 * 4/3

= 2.7 moles

Iron is the limiting reactant

Moles of excess reactant = 3.7 moles - 2.7 moles

= 1 mole

Mass of excess reagent = 1 mole * 18 g/mol

= 18 g

3) 2 moles of octane reacts with 25 moles of oxygen

0.57 mol of octane reacts with 0.57 * 25/2

= 7.125 moles

Oxygen is the reactant in excess

2 moles of octane produces 18 moles of water

0.57 mol of octane produces 0.57 * 18/2

=5.13 moles

Mass of water = 5.13 moles * 18 g/mol

= 92.34 g