Multiple Choice
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?
6. Chemical Quantities & Aqueous Reactions
Solution Stoichiometry
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How many milliliters of 0.325 M HCl are needed to react with 16.2 g of magnesium metal?
2 HCl (aq) + Mg (s) → MgCl2 + H2 (g)
A
33 mL
B
742 mL
C
872 mL
D
3321 mL
E
4100 mL
Verified step by step guidance1
Calculate the moles of magnesium (Mg) using its molar mass. The molar mass of Mg is approximately 24.31 g/mol. Use the formula: \( \text{moles of Mg} = \frac{\text{mass of Mg}}{\text{molar mass of Mg}} \).
Use the stoichiometry of the balanced chemical equation to find the moles of HCl needed. According to the equation, 2 moles of HCl react with 1 mole of Mg. Therefore, multiply the moles of Mg by 2 to find the moles of HCl required.
Use the molarity formula to find the volume of HCl solution needed. The formula is: \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \). Rearrange to solve for volume: \( \text{liters of solution} = \frac{\text{moles of HCl}}{M} \), where M is the molarity of the HCl solution (0.325 M).
Convert the volume from liters to milliliters by multiplying by 1000, since 1 liter = 1000 milliliters.
Compare the calculated volume in milliliters to the given options to determine the correct answer.
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