Answer:
a) Reaction 1: 2K (s) + ZnCl₂ (aq) → 2KCl (aq) + Zn (s)
Reaction 2: C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)
b) 6.02 × 10²³ particles, 1 mole, 136.286 g of ZnCl₂
5 × 6.02 × 10²³ particles or 3.01 × 10²⁴ particles, 5 moles, 5 × 32 or 160 g of O₂
c) One possible molar ratio for each
d) Reaction 1 is a Substitution reaction
Reaction 2 s a combustion reaction
e) The limiting reactant for reaction 1 is ZnCl₂
The limiting reactant for reaction 2 is O₂
f) The excess reactant for reaction 1 is potassium, K
The excess reactant for reaction 2 is butane, C₃H₈
g) Reaction 1: Mass of zinc solid produced is 4.798 grams
Reaction 2: Total mass of gas produced is 12.76 g
h) Reaction 1: Mass of the excess reactant K remaining is 4.26 g
Reaction 2: Mass of the excess reactant C₃H₈ remaining is 9.375 g
Step-by-step explanation:
a) The balanced chemical equation for the reactions are as follows;
Reaction 1: 2K (s) + ZnCl₂ (aq) → 2KCl (aq) + Zn (s)
Reaction 2: C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)
b) 6.02 × 10²³ particles, 1 mole, 136.286 g of ZnCl₂
5 × 6.02 × 10²³ particles or 3.01 × 10²⁴ particles, 5 moles, 5 × 32 or 160 g of O₂
c) Since the equation represents a balanced chemical reaction, only one molar ratio is possible
Similarly, equation 2 represents a balanced chemical equation, therefore, only one molar ratio is possible
d) Reaction 1 reaction, 2K + ZnCl₂ → 2KCl + Zn is a substitution reaction Reaction 2 reaction, C₃H₈ + 5O₂ → 3CO₂ + 4H₂O is a combustion reaction
e) Given 10 g each of K and ZnCl₂, we have
2 moles of K combining with 1 mole of ZnCl₂, to form 2 moles of KCl and 1 mole Zn
Molar mass of K = 39.0983 g/mol
Molar mass of ZnCl₂ = 136.286 g/mol
Number of moles of K = 10/39.0983 = 0.256 moles
Number of moles of ZnCl₂ = 10/136.286 = 0.073 moles
Therefore, 0.073 moles of ZnCl₂ will combine with 2 × 0.073 or 0.147 moles of K, therefore, ZnCl₂ is the limiting reactant
Reaction 2
For reaction 2, we have;
10 g of C₃H₈ and 10 g of O₂
Molar mass of C₃H₈ = 44.1 g/mol
Molar mass of O₂ = 32 g/mol
Therefore number of moles in 10 g of each is given by the following relation
Number of moles of C₃H₈ in 10 g of C₃H₈ = (10 g)/(44.1 g/mol) = 0.227 moles
Number of moles of O₂ in 10 g of O₂ = (10 g)/(32 g/mol) = 0.3125 moles
1 mole of C₃H₈ combining with 5 moles of O₂
Therefore, 0.227 mole of C₃H₈ combines with 5 × 0.227 moles or 1.134 moles of O₂ hence O₂ is the limiting reactant
f) The excess reactant for reaction 1 is potassium, K
The excess reactant in reaction 2 is propane, C₃H₈
g) From the equation of the reaction 2K + ZnCl₂ → 2KCl + Zn, we have 2 moles of KCl aqueous and 1 mole of zinc solid are produced from 1 mole of ZnCl₂ in the reaction, hence 0.073 moles of ZnCl₂ will produce 0.073 moles of Zn
Molar mass of Zn = 65.38 g/mol
Mass of zinc produced = Number of moles × Molar mass
∴ Mass of zinc produced = 0.073 moles × 65.38 g/mol = 4.798 grams
Reaction 2:
In the chemical reaction, C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g), the gaseous products are CO₂ and H₂O. That is 5 moles of O₂, which is the limiting reactant will produce 3 moles of CO₂ and 4 moles of H₂O, hence 0.3125 moles of O₂ will produce 3/5× 0.3125 moles or 0.1875 moles of CO₂ and 4/5×0.3125 moles or 0.25 moles of H₂O
Molar mass of CO₂ = 44.01 g/mol
Molar mass of H₂O = 18.01528 g/mol
Mass of CO₂ produced = Number of moles × Molar mass
∴ Mass of CO₂ produced = 0.1875 moles × 44.01 g/mol = 8.25 g
Similarly mass of H₂O produced = 0.25 moles × 18.01528 g/mol = 4.5 g
Total mass of gas produced = 8.25 + 4.5 = 12.76 g
h)
For Reaction 1
The number of moles of the remaining is presented as follows;
Number of moles of K available = 0.256 moles
Number of moles of K taking part in the reaction = 0.147 moles
Number of moles of K remaining = 0.256 moles - 0.147 moles = 0.109 moles
Mass of K remaining = Number of moles of K × Molar mass of K
Mass of K remaining = 0.109 moles × 39.0983 g/mol = 4.26 g
For reaction 2
The number of moles of the remaining is presented as follows;
0.3125 moles of of O₂ will combine with 1/5×0.227×0.3125 moles or 0.0142 moles of C₃H₈,
Number of moles of C₃H₈ available = 0.227 moles
Number of moles of C₃H₈ taking part in the reaction = 0.0142 moles
Number of moles of C₃H₈ remaining = 0.227 moles - 0.0142 moles = 0.213 moles
Mass of C₃H₈ remaining = Number of moles of C₃H₈ × Molar mass of C₃H₈
Mass of C₃H₈ remaining = 0.213 moles × 44.1 g/mol = 9.375 g.