How many grams of PbBr2 will precipitate when excess CuBr2 solution is added to 68.0 mL of 0.682 M Pb(NO3)2 solution?
回答 (2)
2016-02-24 6:58 am
CuBr₂(aq) + Pb(NO₃)₂(aq) → PbBr₂(aq) + Cu(NO₃)₂(aq)
1 mole of Pb(NO₃)₂ gives 1 mole of PbBr₂ precipitate.
No. of moles of Pb(NO₃)₂ used = 0.682 × (68.0/1000) = 0.0464 mol
No. of moles of PbBr₂ precipitate = 0.0464 mol
Molar mass of PbBr₂ precipitate = 207.2 + 79.9×2 = 367 g/mol
Mass of PbBr₂ precipitate = 0.0464 × 367 = 17.0 g
1 mole of Pb(NO₃)₂ gives 1 mole of PbBr₂ precipitate.
No. of moles of Pb(NO₃)₂ used = 0.682 × (68.0/1000) = 0.0464 mol
No. of moles of PbBr₂ precipitate = 0.0464 mol
Molar mass of PbBr₂ precipitate = 207.2 + 79.9×2 = 367 g/mol
Mass of PbBr₂ precipitate = 0.0464 × 367 = 17.0 g
2016-02-24 6:49 am
367 g/mole X 0.682 mole/L X 0.068 L = 17.02 g
收錄日期: 2021-04-18 14:31:01
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