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angelinaissoocp33868
08.04.2020 •
Chemistry
4. When aqueous solutions of lead(II) ion are treated with potassium chromate solution, a bright yellow precipitate of lead(II) chromate, PbCrO4, forms. How many grams o lead chromate form when a 1.00-g sample of Pb(NO3)2 is added to 25.0mL of 1.00M K2CrO4 solution
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Ответ:
Lead chromate form = 0.97 g
Explanation:
The Balanced chemical equation of the given reaction is as follows
K₂CrO₄ + Pb(NO₃)₂ → PbCrO₄ + 2 KNO₃
Given that
Mass of Pb(NO₃)₂ = 1 g
Mole of 25 ml of 1.00 (M) K₂CrO₄
Mole = 25 x 1 milimol = 25 x 10⁻³
Using mole ratio method to find limiting reagent
Hence K₂CrO₄(potassium chromate) is a limiting reagent.
So lead chromate formed = mole X molar mass
= 0.003 X 323.19 g
= 0.97 g
∴ 0.97 g lead chromate form when a 1.00-g sample of Pb(NO₃)₂ is added to 25.0 ml of 1.00 M K₂CrO₄ solution.
Ответ: