50 g KCIO3 dissolved in 200 mL H20 at 85 C Question 2 0.1003 M Pb(NO3)z. Work Te

Question

50 g KCIO3 dissolved in 200 mL H20 at 85 C Question 2 0.1003 M Pb(NO3)z. Work Temperature ("C 124/solubility htm -Attempt this question without getting help, using only a periodic table. e volume of 0.06478 M NaCI that is required to completely react with 24.33 mL of 2 NaCI(aq) +Pb(NO3)2(aq) PbCl2(s) +2NaNO3(aq) Answer ?looks great-keep it missing some instructions review molarity calculations only minor errors-double Instructor Feedback/Recommendations upl watch sig figs check details of wording watch units missing points on graph missing/incorrect mol-mol r review definitions need more practicecorrect setup; check calcs again of solutions ratio check details see instructor or tutor for better explanation of topic reading graph

Explanation / Answer

2.

2NaCl + Pb(NO3)2 --> PbCl2 + 2NaNO3

moles Pb(NO3)2 present = 0.1003 M x 24.33 ml = 2.4403 mmol

moles NaCl required = 2 x 2.4403 = 4.8806 mmol

volume NaCl solution needed = 4.8806 mmol/0.06478 M = 75.34 ml

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