A solution of potassium chromate reacts with a solution of lead II nitrate to produce a yellow precipitate of lead II chromate and a solution of potassium nitrate. How many grams of potassium nitrate can be produced from 15 grams of potassium chromate?
Added by Diane J.
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Given: Molar mass of K2CrO4 = 294.2 g/mol Mass of K2CrO4 = 15 g Number of moles of K2CrO4 = Mass/Molar mass Number of moles of K2CrO4 = 15 g / 294.2 g/mol Number of moles of K2CrO4 ≈ 0.051 moles Show more…
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A solution of potassium chromate reacts with a solution of lead(II) nitrate to produce a yellow precipitate of lead(II) chromate and a solution of potassium nitrate. \begin{equation} \begin{array}{l}{\text { a. Write the balanced chemical equation. }} \\ {\text { b. Starting with } 0.250 \text { mol of potassium chromate, }} \\ {\text { determine the mass of lead chromate formed. }}\end{array} \end{equation}
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 of lead chromate form when a 2.52 g sample of Pb(NO3)2 is added to 29.8 mL of 3.99 M K2CrO4 solution?
Oluwapelumi K.
When aqueous solutions of lead(II) ion are treated with potassium chromate solution, a bright yellow precipitate of lead(II) chromate, $\mathrm{PbCrO}_{4}$ , forms. How many grams of lead chromate form when a $1.00-\mathrm{g}$ sample of $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ is added to 25.0 $\mathrm{mL}$ of 1.00 $\mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}$ solution?
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