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Potassium iodide reacts with lead(II) nitrate in the following precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) What minimum volume of 0.400 M potassium iodide solution is required to completely precipitate all of the lead in 310.0 mL of a 0.112 M lead(II) nitrate solution?

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Answer:

0.1736 L or 173.6 ml

Explanation:

Number of moles of lead II nitrate is obtained by;

Number of moles = concentration × volume of solution

Concentration= 0.112 M

Volume of solution= 310 ml

n= 0.112 × 310/1000

n= 0.03472 moles

From the reaction equation;

2 moles of potassium iodide reacted with 1 mole of lead II nitrate

x moles of potassium iodide will react with 0.03472 moles of lead II nitrate

x= 2 × 0.03472 moles= 0.06944 moles of potassium iodide

Volume of potassium iodide solution = number of moles/ concentration = 0.06944/ 0.4

Volume of potassium iodide solution= 0.1736 L or 173.6 ml

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