Suppose we have a solution of lead nitrate, Pb(NO3)2(aq). A solution of NaCl(aq) is added slowly until no furt?
2016-06-13 10:09 am
Suppose we have a solution of lead nitrate, Pb(NO3)2(aq). A solution of NaCl(aq) is added slowly until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 10.03 g of PbCl2(s) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) solution.
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2016-06-18 1:36 pm
✔ 最佳答案
(10.03 g PbCl2) / (278.1164 g PbCl2/mol) x (1 mol Pb(NO3)2 / 1 mol PbCl2) / (0.2000 L) = 0.1803 mol Pb(NO3)2/L
2016-06-13 1:07 pm
Molar mass of PbCl₂ = (207.2 + 35.45×2) g/mol = 278.1 g/mol
Pb(NO₃)₂(aq) + 2NaCl(aq) → PbCl₂(s) + 2NaNO₃(aq)
No. of moles of PbCl₂ formed = (10.03 g) / (278.1 g/mol) = 0.03607 mol
No. of moles of Pb(NO₃)₂ reacted = (0.03607 mol) × (1/1) = 0.03607 mol
Molarity of Pb(NO₃)₂ solution = (0.03607 mol) / (200.0/1000 L) = 0.1804 M
Pb(NO₃)₂(aq) + 2NaCl(aq) → PbCl₂(s) + 2NaNO₃(aq)
No. of moles of PbCl₂ formed = (10.03 g) / (278.1 g/mol) = 0.03607 mol
No. of moles of Pb(NO₃)₂ reacted = (0.03607 mol) × (1/1) = 0.03607 mol
Molarity of Pb(NO₃)₂ solution = (0.03607 mol) / (200.0/1000 L) = 0.1804 M
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