F.4 Chemistry 兩條(15分****)

answer:

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第一條:
(a)no. of moles of CO2
= 14/(12+2×16)
= 0.318 mol
no. of moles of KO2 needed
= 0.318 mol
mass of KO2 needed
= (0.318)(39+2×16)
= 22.59
(b)mass needed
= 22.59÷0.8
= 28.2 g


第二條:

(a) Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)

(b) No. of moles of Cu used
= (4.36﹣2.21)/63.5
= 0.03386 mol
No. of moles of Ag formed
= 2×No. of moles of Cu used
= 2×0.03386 mol
= 0.06772 mol
Mass of Ag formed
= 0.06772×108
= 7.3134 g