Chemistry help needed badly?

[匿名]

2017-07-25 10:51 pm

Studying for a chemistry exam and I can't come up with the right answer for these questions. Thanks.

1. Consider the following balanced chemical reaction:
2 HCl(aq) + Pb(OH)2(s) -> 2 H2O(l) + PbCl2(s)
In a laboratory, a student mixes 35.00 mL of 2.200 M aqueous HCl with 10.00 g of solid Pb(OH)2. How many grams of PbCl2 will be produced in this reaction?

2. Consider the following double displacement or metathesis reaction of Na3PO4(aq) with MgCl2(aq).

A. Write a balanced chemical equation including the states of matter for each species.
B. Write a balanced ionic equation.
C. Write a balanced net ionic equation.
D. What are the spectator ions?

回答 (1)

wanszeto

2017-07-25 11:20 pm

✔ 最佳答案

1.
Molar mass of Pb(OH)₂ = (207.2 + 16.0×2 + 1.0×2) g/mol = 241.2 g/mol

Initial number of moles of HCl = (2.200 mol/L) × (35.00/1000 L) = 0.07700 mol
Initial number of moles of Pb(OH)₂ = (10.00 g) / (241.2 g/mol) = 0.04146 mol

2HCl(aq) + Pb(OH)₂(s) → 2H₂O(l) + PbCl₂(s)
Mole ratio HCl : Pb(OH)₂ = 2 : 1

If HCl completely reacts :
Number of moles of Pb(OH)₂ needed = (0.07700 mol) × (1/2) = 0.03850 mol < 0.04146 mol
Hence, Pb(OH)₂ is in excess, and HCl completely reacts.

According to the equation, mole ratio HCl : PbCl₂ = 2 : 1
Number of moles of HCl reacted = 0.07700 mol
Number of moles of PbCl₂ produced = (0.07700 mol) × (1/2) = 0.03850 mol

Molar mass of PbCl₂ = (207.2 + 35.5×2) g/mol = 278.2 g/mol
Mass of PbCl₂ produced = (0.03850 mol) × (278.2 g/mol) = 10.71 g

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2.
A.
2Na₃PO₄(aq) + 3MgCl₂(aq) → Mg₃(PO₄)₂(s) + 6NaCl(aq)

B.
6Na⁺(aq) + 2PO₄³⁻(aq) + 3Mg²⁺(aq) + 6Cl⁻(aq) → Mg₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

C.
3Mg²⁺(aq) + 2PO₄³⁻(aq) → Mg₃(PO₄)₂(s)

D.
The spectator ions are Na⁺(aq) and Cl⁻(aq) ions.

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