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1. Consider Reactions 3 and 14 studied in this lab. Write the balanced molecular equation (identical to what you completed in the previous section), the complete ionic equation and the net ionic equation for these reactions. Include all physical states, and circle the spectator ions in the complete ionic equations.
Reaction 3: Aqueous sodium phosphate + aqueous copper(II) sulfate
Balanced Molecular Equation
(from page 1): Complete
Ionic Equation:
Net Ionic Equation:
Reaction 14: Aqueous iron(III) chloride + aqueous ammonium hydroxide
Balanced Molecular Equation
(from page 3): Complete
Ionic Equation:
Net Ionic Equation:
2. Predict the products for the following single and double displacement reactions, and write balanced molecular equations including physical states) for each of them. If you predict that no reaction will occur, write "NR", followed by a brief explanation.
A. Aluminum metal + aqueous silver acetate
B. Aqueous zinc nitrate + aqueous lithium chloride
C. Hydrobromic acid + solid magnesium sulfite
D. Aqueous rubidium hydroxide + perchloric acid
E. Tin metal + phosphoric acid
F. Aqueous lithium chromate + aqueous gold(111) iodide

Respuesta :

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Answer:

See explanation

Explanation:

For reaction 3;

Molecular equation

2Na3PO4(aq) + 3CuSO4(aq) -----> 3Na2SO4(aq) + Cu3(PO4)2(s)

Complete ionic equation;

6Na^+(aq) + 2PO4^3-(aq) + 3Cu^2+(aq) + 3SO4^2-(aq) ------> 6Na^+(aq) + 3SO4^2-(aq) + Cu3(PO4)2(s)

Net ionic equation;

3Cu^2+(aq) + 2PO4^3-(aq)  -----> Cu3(PO4)2(s)

Reaction 14

Molecular equation

2FeCl3(aq) + 6NH4OH(aq) ------>2Fe(OH)3(s) + 6NH4Cl

Complete ionic equation:

2Fe^2+(aq) + 6Cl^-(aq) + 6NH4^+(aq) + 6OH^-(aq) -----> 2Fe(OH)3(s) + 6NH4^+(aq) +  6Cl^-(aq)

Net ionic equation;

2Fe^2+(aq) +  6OH^-(aq)  -----> 2Fe(OH)3(s)

2.

Al(s) + 3Ag(CH3COO)(aq) ----->Al(CH3COO)3(aq) + 3Ag(s)

Zn(NO3)2(aq) +2LiCl(aq) ------> 2LiNO3(aq) + ZnCl2(aq)

2HBr(aq) + MgSO3(s) ------>MgBr2(aq) + H2O(l) + SO2(g)

RbOH(aq) + HClO4(aq) -----> RbClO4(aq) + H2O(l)

3Sn(s) + 4H3PO4(aq)----->Sn3(PO4)4(aq) + 6H2(g)

3Li2CrO4(aq) + 2AuI3(aq) -------> 6LiI(aq) + Au2(CrO4)3(s)

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