Chapter 8

Question 1

What is primary gas that causes global warming?

Answer 1

CO₂

Question 2

Gasoline is C₈H₁₈(l). Write equation for its combustion in CO₂(g) and H₂O(g)

Answer 2

2 C₈H₁₈(l) + 25 O₂(g) –> 16 CO₂(g) + 18 H₂O(g)

Question 3

The numerical relationship between the chemical quantities in a blanced chemical equation is called?

Answer 3

reaction stoichiometry

Question 4

In the reaction 3 H₂(g) + N₂(g) –> 2NH₃(g) what is the molar ratio?

Answer 4

3 mol H₂: 1 mol N₂: 2 mol NH₃

Question 5

In the reaction 3 H₂(g) + N₂(g) –> 2NH₃(g) if 9 moles of H2(g) reacts how many moles of NH₃(g) are produced?

Answer 5

6

Question 6

In the reaction 2 Na(s) + Cl₂(g) –> 2NaCl(s), how many moles of NaCl are produced with 3.4 moles of Cl₂(g)?

Answer 6

6.8 moles

Question 7

In the reaction 2 C₈H₁₈(l) + 25 O₂(g) –> 16 CO₂(g) + 18 H₂O(g) which consumes 500 g of C₈H₁₈(l), how much CO₂(g) is produced?

Answer 7

1500 g CO₂(g)

since

1 mole C₈H₁₈ = 114.3 g

2 moles C₈H₁₈ –> 16 moles CO₂

1 mole CO₂ = 44.01 g

Question 8

In the reaction 6 CO₂(g) + 6 H₂O(l) –> 6 O₂(g) + C₆H₁₂O₆(aq), how many grams of glucose can be obtained from 58.5 g of CO₂?

Answer 8

39.9 g C₆H₁₂O₆

since

1 mol CO₂ = 44.01 g

6 mol CO2 –> 1 mol C₆H₁₂O₆

1 mol C₆H₁₂O₆ = 180.2 g

Question 9

In the reaction 4 NO₂(g) + O₂(g) + 2 H₂O(l) –> 4 HNO₃(aq), how much HNO₃ is proudced for 1.5 x 10³ kg of NO₂?

Answer 9

2.1 x 10³ kg HNO₃

since

1000 g = 1 kg

1 mol NO₂ = 46.01 g

4 mol NO2 –> 4 mol HNO₃

1 mol HNO₃ = 63.02 g

Question 10

In the reaction Ti(s) + 2 Cl₂(g) –> TiCl₄(s), if there are 1.8 mmol of Ti and 3.2 mol of Cl what is limiting reactant and the theoretical yield of TiCl₄?

Answer 10

stoichiometric ratio

1 Ti : 2 Cl₂ : 1 TiCl₄

1. 8 mol Ti –> 1.8 mol TiCl₄
2. 2 mol Cl₂ –> 1.6 mol TiCl₄

so limiting ractant is CL₂ and theoretical yield is 1.6 mol TiCl₄

Question 11

In the reaction 2 Al(s) + 3 Cl₂(g) –> 2 AlCl₂(s) with 0.552 mol of Al and 0.887 mol of Cl₂ what is limiting reactant and theoretical yield of AlCl₂?

Answer 11

stoichiometric relation

2 mol Al : 3 mol Cl2 : 2 mol AlCl₂

0. 552 mol Al –> 0.552 AlCl₂
1. 887 mol Cl₂ –> 0.591 mol AlCl₂

limiting reactant is Al; theoetical yield of ALCL₂ is 0.552 mol,

Question 12

In the reaction 2 Na(s) + F₂(g) –> 2 NaF(s), if there are 4.8 mol of Na and 2.6 mol of F₂ what is limiting reactant and theoretical yield of NaF?

Answer 12

stoichiometric ratio

2 Na : 1 F₂ : 2 NaF

4. 8 mol Na –> 4.8 mol NaF
5. 6 mol F₂ –> 5.2 mol NaF

Na is limiting reactant and theoretical yield is 4.8 mol NaF.

Question 13

In the reaction 2 NO(g) + 5 H₂(g) –> 2 NH₃(g) + 2 H₂O(g) what is maximum amount of ammonia in grams that can be formed from 45.8 g NO and 12.4 g H₂?

Answer 13

Molar mass: NO 30.01 g/mol, H₂ 2.02 g/mol, NH₃ 17.04 g/mol

Stoichiometric ratio: 2 NO : 5 H₂ : 2 NH₃ : 2H₂O

45. 8 g NO –> 26.0 g NH₃
46. 4 g H₂ –> 41.8 g NH₃

NO is limiting ractant and only 26.0 g NH₃ can be formed

Question 14

In the reaction Cu₂O(s) + C(s) –> 2 Cu(s) + CO(g) when 11.5 g of C racts with 114.5 g Cu₂O, 87.4 g of Cu are obtained. What is the limiting reactant, theoretical yield and percent yield?

Answer 14

Molar mass: C 12.01 g/mol, Cu₂O 143.10 g/mol, Cu 63.55 g/mol

Stoichiometric ratio: 1 Cu₂O : 1 C : 2 Cu : 1 CO

11. 5 g C –> 122 g Cu
12. 5 g Cu₂O –> 101.7 g Cu

Cu₂O limiting reactant with a theoretical yield of 101.7 g

percent yield = 87.4/101.7 = 85.9%

Question 15

In the reaction C₃H₈(g) + 5 O₂(g) –> 3 CO₂(g) + 4 H₂O(g) H_rxn = -2044 kJ assume 1.18 x 10⁴ g of C₃H₈ are burned how much heat is released?

Answer 15

Molar mass C₃H₈ 44.11 g/mol

1.18 x 10⁴ g C₃H₈ = 268 mol ofC₃H₈

Heat released = 268 * 2044kJ = 5.47 x 10⁵ kJ