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28.0 x 0.6424 = 18
25.0 x .6424 = 16.1
Mass = 34.1
Answer:
The question is incomplete and confusing.
- In the complete ionic equation you write all the ions that are formed. Those are: Pb²⁺, NO₃⁻, K⁺, and I⁻. They all are present in the complete ionic equation.
- In the net ionic equation, the spectator ions do not appear. They are: NO₃⁻ and K⁺. They would not be present in the net ionic equation, but they do in the complete ionic equation.
See below the details.
Explanation:
Which compound will not form ions?
<u />
<u>1. Write the balanced molecular equation:</u>
- Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)
<u />
<u>2. Write the ionizations for the ionic aqueous compounds:</u>
<u />
- Pb(NO₃)₂(aq) → Pb⁺²(aq) + 2NO₃⁻(aq)
- 2KI(aq) → 2K⁺(aq) + 2I⁻(aq)
- 2KNO₃(aq) → 2K⁺(aq) + 2NO₃⁻(aq)
<u />
<u>3. Write the complete ionic equation:</u>
Pb⁺²(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq)
Hence, since PbI₂(s) does not ionize, but stays in solid form, it will not form ions.
All, Pb⁺², NO₃⁻, K⁺, and I⁻ will be present in the total ionic equation.
It is in the net ionic equation that the spectator ions are removed. Those, are NO₃⁻ and K⁺, because they are on both sides of the complete ionic equation.
Answer:
4.0473g. Attached in image are the calculations I used to reach this answer.
