FeBr3(S) + 3 AgNO3(aq)--> 3AgBr(s) + Fe(NO3)3 (aq)
help!! please explain how you got the answer!
what mass of AgBr, silver bromide, is produced by reaction of 88.8 g of FeBr3, with excess AgNO3?
2017-11-06 12:03 pm
回答 (1)
2017-11-06 12:14 pm
Molar mass of FeBr₃ = (55.8 + 79.9×3) g/mol = 295.5 g/mol
Molar mass of AgBr = (107.9 + 79.9) g/mol = 187.9 g/mol
(The molar masses used in your lessons should be used instead.)
3AgNO₃(aq) + FeBr₃(aq) → 3AgBr(s) + Fe(NO₃)₃(aq)
Mole ratio FeBr₃ : AgBr = 1 : 3
No. of moles of FeBr₃ reacted = (88.8 g) / (295.5 g/mol) = 0.3 mol
No. of moles of AgBr produced = (0.3005 mol) × 3 = 0.9 mol
Mass of AgBr produced = (0.9 mol) × (187.9 g/mol) = 169 g
Molar mass of AgBr = (107.9 + 79.9) g/mol = 187.9 g/mol
(The molar masses used in your lessons should be used instead.)
3AgNO₃(aq) + FeBr₃(aq) → 3AgBr(s) + Fe(NO₃)₃(aq)
Mole ratio FeBr₃ : AgBr = 1 : 3
No. of moles of FeBr₃ reacted = (88.8 g) / (295.5 g/mol) = 0.3 mol
No. of moles of AgBr produced = (0.3005 mol) × 3 = 0.9 mol
Mass of AgBr produced = (0.9 mol) × (187.9 g/mol) = 169 g
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