CHEMISTRY question please help! Just need you to check my answer. THANKS?

👨🏻‍🏫 學術台化學

[匿名]

2017-10-25 10:48 am

Potassium iodide reacts with lead(II) nitrate in the following precipitation reaction:
2KI(aq)+Pb(NO3)2(aq)→2KNO3(aq)+PbI2(s)

The answer is 378 mL I just need to know how that is solved.

更新1:

What minimum volume of 0.200 M potassium iodide solution is required to completely precipitate all of the lead in 195.0 mL of a 0.194 M lead(II) nitrate solution?

回答 (1)

戇戇居士

2017-10-25 11:14 am

✔ 最佳答案

2KI(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbI₂(s)
Mole ratio KI : Pb(NO₃)₂ = 2 : 1

No. of moles of Pb(NO₃)₂ reacted = (0.194 mmol/mL) × (195.0 mL) = 37.83 mmol
Minimum no. of moles of KI needed = (37.83 mmol) × 2 = 75.66 mmol
Volume of 0.200 M KI needed = (75.66 mmol) / (0.2 mmol/mL) = 378 mL (to 3 sig. fig.)

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