When 45 mL of 1.3 M sr(no3)2 mixes with excess alcl3 how many grams of srcl2 are produced?
回答 (2)
2018-04-11 10:19 am
3Sr(NO₃)₂ + 2AlCl₃ → 3SrCl₂ + 2Al(NO₃)₃
Mole ratio Sr(NO₃)₂ : SrCl₂ = 3 : 3 = 1 : 1
No. of moles of Sr(NO₃)₂ reacted = (1.3 mol/L) × (45/1000 L) = 0.0585 mol
No. of moles of SrCl₂ produced = 0.0585 mol
Molar mass of SrCl₂ = (87.62 + 35.45×2) g/mol = 158.52 g/mol
Mass of SrCl₂ produced = (0.0585 mol) × (158.52 g/mol) = 9.27342 g ≈ 9.3 g (to 2 sig. fig.)
Mole ratio Sr(NO₃)₂ : SrCl₂ = 3 : 3 = 1 : 1
No. of moles of Sr(NO₃)₂ reacted = (1.3 mol/L) × (45/1000 L) = 0.0585 mol
No. of moles of SrCl₂ produced = 0.0585 mol
Molar mass of SrCl₂ = (87.62 + 35.45×2) g/mol = 158.52 g/mol
Mass of SrCl₂ produced = (0.0585 mol) × (158.52 g/mol) = 9.27342 g ≈ 9.3 g (to 2 sig. fig.)
2018-04-11 10:53 am
None -- everything is water soluble.
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