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Two chemical reaction occur:

#1. Solid potassium and aqueous solution zinc chloride yields zinc solid and potassium
chloride (aq)
#2. A hydrocarbon gas, C 3 H 8 reacts with oxygen to form water and carbon dioxide

For each reaction (#1,#2) answer the following questions and SHOW your WORK for
credit. Answers alone will only get a 60%... Remember to use significant figures, units.
a) What is the balanced chemical equation for reaction #1?
What is the balanced chemical equation for reaction #2?
b) Express zinc chloride as particles, moles, mass in reaction #1
Express the oxygen as particles, moles, mass in reaction #2
c) How many molar ratios are possible in the chemical equation #1
How many molar ratios are possible in the chemical equation #2
d) What is the type of chemical reaction for reaction #1?
What is the type of chemical reaction for reaction #2?
If you are given 10.0 g of each reactant in the chemical reactions
e) What is the limiting reactant of reaction #1?
What is the limiting reactant of reaction #2?
f) What is the excess reactant in reaction #1?
What is the excess reactant in reaction #2
g) How much of the solid is produced in reaction #1
How much of the gas is produced in reaction #2
h) How much of the excess reactant remains after the #1 reaction?
How much of the excess reactant remains after the #2 reaction?

User Roadkillnz
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1 Answer

4 votes

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.

User Shea
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