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andrise andrise
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8 years ago
What is the mass of silver bromide (187.77 g/mol) precipitated from 2.96 g of iron(III) bromide (295.55 g/mol)?

 __FeBr3(s) + __AgNO3(aq) → __AgBr(s) + __Fe(NO3)3(aq)
      
1.88 g
      
5.64 g
      
0.627 g
      
3.76 g
      
0.940 g
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Educator
8 years ago
FeBr3 + 3 Ag{+} → 3 AgBr + Fe{3+}

(2.96 g FeBr3) / (295.55 g FeBr3/mol) x (3 mol AgBr / 1 mol FeBr3) x
(187.77 g AgBr/mol) = 5.64 g AgBr
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