Exam 4 Chem

A. This is a Balanced Equation

B. This is a Single Replacement reaction

C. 5.38g of Zn = .10 moles Zn (rounded to 2 sig figs)

D. 7.28g CuSO₄ = .10 moles CuSO₄ (rounded to 2 sig figs)

E. If you react 5.38g of Zn with 7.28g CuSO4, than Zn is the limiting reagent

F. If you react 5.38g of Zn with 7.28g CuSO₄, you can produce .0456 moles of ZnSO₄ (rounded to 3 sig figs)

G. If you react 5.38g of Zn with 7.28g CuSO₄, you can produce 9.23g of ZnSO₄ (rounded to 3 sig figs)

H. The Cuprous Sulfate in the above reaction is dissolved in water

I. Since the change in enthalpy for the reaction above is –217kJ/mol, than it is an endothermic reaction.

J. Since there is no significant decrease in entropy, the reaction above would tend to move to completion.

A. This is a Balanced Equation

B. This is a Double replacement reaction

C. 3.2g of Methane = .20 moles Methane (rounded to 2 sig figs)

D. 9.6g O₂ = .30 moles Oxygen gas (rounded to 2 sig figs)

E. If you react 3.2g of Methane with 9.6g Oxygen gas, than Methane is the limiting reagent

F. If you react 3.2g of Methane with 9.6g Oxygen gas, you can produce .20 moles of Carbon Dioxide (rounded to 2 sig figs)

G. If you react 3.2g of Methane with 9.6g Oxygen gas, you can produce 6.6 g of Carbon Dioxide (rounded to 2 sig figs)

H. The water in the above reaction is a gas

I. Since the change in enthalpy for the reaction above is –890kJ/mol, than it is an exothermic reaction.

J. From the number of gas particles in the reaction, we can see that a large increase in entropy drives the reaction toward completion.

A. This is a Balanced Equation

B. This is a Double replacement reaction

C. 24g of MgI₂ = 0.30 moles MgI₂ (rounded to 2 sig figs)

D. 12g of NaOH = 0.30 moles NaOH (rounded to 2 sig figs)

E. If you react 24g of MgI₂ with 12g of NaOH, than Sodium Hydroxide is the limiting reagent

F. If you react 24g of MgI₂ with 12g of NaOH, than you can produce 0.086 moles of Magnesium Hydroxide (rounded to 2 sig figs)

G. If you react 24g of MgI₂ with 12g of NaOH, than you can produce 5.0 g of Magnesium Hydroxide (rounded to 2 sig figs)

H. The Magnesium Hydroxide in this solution is a solid that will fall out of solution

I. Since the change in enthalpy for the reaction above is negative, than it is an endothermic reaction.

J. The Precipitation of Magnesium Hydroxide will reduce entropy and could drive it toward completion.

A. This is a Balanced Equation

B. This is a Synthesis reaction

C. 1.2g of H₂ = 0.60 moles H₂ (rounded to 2 sig figs)

D. 2.4g of N₂ = 0.30 moles N₂ (rounded to 2 sig figs)

E. If you react 3.3 L of H₂ with 3.3 L of N₂ than Nitrogen is the limiting reagent

F. If you react 3.3 L of H₂ with 3.3 L of N₂ than you can produce 2.2 of Ammonia (rounded to 2 sig figs)

G. If you react 3.3 L of H₂ with 3.3 L of N₂ than you can produce 0.098 moles of Ammonia (rounded to 2 sig figs)

H. This reaction increases entropy

I. If the change in enthalpy is -96kJ/mol, this reaction is endothermic

J. This is an exothermic reaction which decreases entropy, which means the products and reactants will find equilibrium

A. This is a Balanced Equation

B. This is a Neutralization reaction

C. 1.28g of HI = 0.0010 moles HI (rounded to 2 sig figs)

D. 1.13g of KOH = 0.0020 moles N₂ (rounded to 2 sig figs)

E. If you react 1.28g of HI with 1.13g of KOH than KOH is the limiting reagent

F. If you react 1.28g of HI with 1.13g of KOH than you can produce .001 moles of Potassium Iodide (rounded to 2 sig figs)

G. If you react 1.28g of HI with 1.13g of KOH than you can produce 1.7 g of of Potassium Iodide (rounded to 2 sig figs)

H. In this reaction, the HI is an acid.

I. In this reaction, the KOH is a base.

J. This is an exothermic reaction, which means the change in enthalpy is positive.