15.0 mL of a 0.50 M Pb(NO32 solution is mixed with 15.0 mL of a 0.50 M NaCl solution. The following reaction goes to completion. Pb(NO3)2(aq) + 2NACI(aq) → PbCh(s) + 2NANO3(aq) 1. What mass of precipitate is expected to form? grams (2 sig figs) 2. What ion is limiting? (Lead Il or Chloride) 2. What is the volume of the resulting solution? (assume the volumes are additive) mL (1 decimal place)

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15.0 mL of a 0.50 M Pb(NO3)2 solution is mixed with 15.0 mL of a 0.50 M NaCl solution. The following reaction goes to completion.
Pb(NO3)2(aq) + 2NACI(aq) → PbCh(s) + 2NANO3(aq)
1. What mass of precipitate is expected to form?
grams (2 sig figs)
2. What ion is limiting?
(Lead II or Chloride)
2. What is the volume of the resulting solution? (assume the volumes are additive)
mL (1 decimal
place)
3. What is the new molarity of nitrate ions in the resulting solution?
M (2 sig figs)
4. What is the new molarity of sodium ions in the resulting solution?
M (2 sig figs)
4. What is the molarity of the excess reactant ions?
M (2 sig figs)
Transcribed Image Text:15.0 mL of a 0.50 M Pb(NO3)2 solution is mixed with 15.0 mL of a 0.50 M NaCl solution. The following reaction goes to completion. Pb(NO3)2(aq) + 2NACI(aq) → PbCh(s) + 2NANO3(aq) 1. What mass of precipitate is expected to form? grams (2 sig figs) 2. What ion is limiting? (Lead II or Chloride) 2. What is the volume of the resulting solution? (assume the volumes are additive) mL (1 decimal place) 3. What is the new molarity of nitrate ions in the resulting solution? M (2 sig figs) 4. What is the new molarity of sodium ions in the resulting solution? M (2 sig figs) 4. What is the molarity of the excess reactant ions? M (2 sig figs)
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