Answer:
66.7%
Explanation:
The reaction for the titration of the excess ferrous ion is:
- 5Fe⁺² + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O
We calculate the moles of Fe⁺² from the used moles of KMnO₄:
- 0.02 M * 15.0 mL = 0.30 mmol KMnO₄
- 0.3 mmol KMnO₄ *
= 1.5 mmol Fe⁺²
Then we substract those 0.30 mmol from the original amount used:
- 0.1 M * 50.0 mL = 5.0 mmol Fe⁺²
- 5.0 - 1.5 = 3.5 mmol Fe⁺²
The reaction between ferrous ammonium sulfate and MnO₂ is:
- 2Fe⁺² + MnO₂ + 4H⁺ → 2Fe³⁺ + Mn²⁺ + 2H₂O
So we convert those 3.5 mmol Fe⁺² that were used in this reaction to MnO₂ moles:
- 3.5 mmol Fe⁺² *
= 1.75 mmol MnO₂
Then we convert MnO₂ to Mn₃O₄, using the reaction:
- 1.75 mmol MnO₂ *
= 0.583 mmol Mn₃O₄
Finally we convert Mn₃O₄ moles to grams:
- 0.583 mmol Mn₃O₄ * 228.82 mg/mmol = 133.40 mg Mn₃O₄
And calculate the percent
- 133.40 / 200 * 100% = 66.7%
Incomplete question. The full question read;
You are analyzing water that is known to contain silver nitrate, AgNO3. You decide to determine the amount of silver nitrate using gravimetric analysis based on the reaction:
Ag+ + Cl– → AgCl
You add excess NaCl to a 100 ml sample of the water and find that 1.2 g of AgCl solid forms.
If you added excess NaCl to a 200 ml sample of water from the same source, how many grams of AgCl solid would you expect to form?
Answer:
<u>0.6 gram</u>
Explanation:
<em>Remember</em>, we were first told when NaCl is added to a 100 ml sample of the water it results in the formation of 1.2 g of AgCl.
Hence, if the volume of water is increased 2x to 200 ml from 100 ml, and NaCl is added to it, then the expected number of grams should be 0.6 (1.2g/2). That is, with increased volume, the amount of dissolution of AgCl is increased.
Explanation:
The reaction is given as;
Pb ( N O ₃ ) ₂ (aq) + K I (aq) → Pb I ₂ + K N O ₃
To ensure the reaction is balanced, there has to be equal number of toms of each element present in both the reactant and product side of the reaction.
This leads us to;
Pb ( N O ₃ ) ₂ (aq) + 2 K I (aq) → Pb I ₂(s) + 2 K N O ₃(aq)
Ths way all atoms of elements are the same.
