Multiple Choice
How many milliliters of 0.534 M H₂SO₄ are required to neutralize 245 mL of 0.633 M KOH?
6. Chemical Quantities & Aqueous Reactions
Solution Stoichiometry
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What minimum volume of 0.248 M potassium iodide solution is required to completely precipitate all of the lead in 190.0 mL of a 0.100 M lead(II) nitrate solution?
A
152.0 mL
B
38.3 mL
C
76.6 mL
D
95.0 mL
Verified step by step guidance1
Write the balanced chemical equation for the reaction: \( \text{Pb(NO}_3\text{)}_2 (aq) + 2 \text{KI} (aq) \rightarrow \text{PbI}_2 (s) + 2 \text{KNO}_3 (aq) \). This shows that 1 mole of lead(II) nitrate reacts with 2 moles of potassium iodide.
Calculate the moles of lead(II) nitrate in the solution using the formula: \( \text{moles} = \text{concentration} \times \text{volume} \). Here, the concentration is 0.100 M and the volume is 190.0 mL (convert this to liters by dividing by 1000).
Using the stoichiometry from the balanced equation, determine the moles of potassium iodide required. Since 1 mole of lead(II) nitrate reacts with 2 moles of potassium iodide, multiply the moles of lead(II) nitrate by 2 to find the moles of potassium iodide needed.
Calculate the volume of 0.248 M potassium iodide solution required using the formula: \( \text{volume} = \frac{\text{moles}}{\text{concentration}} \). Use the moles of potassium iodide calculated in the previous step and the given concentration of 0.248 M.
Convert the volume from liters to milliliters by multiplying by 1000, as the final answer should be in milliliters.
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