WLv2 | Online teaching x m/lrn/takeAssignment/takeCovalentActivity do?locator as
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WLv2 | Online teaching x m/lrn/takeAssignment/takeCovalentActivity do?locator assignment-take&takeAssignment; SessionL ocator assignment-take Use the Relerences to access important values if needed for this question. According to the following reaction, how many grams of iron(II) nitrate will be formed upon the complete reaction of 26.6 grams of silver nitrate with excess iron(II) chloride? iron(II) chloride (aq) silver nitrate (aq)ron(II) nitrate (aq) + silver chloride (s) grams iron(II) nitrate Submit Answer Try Another Version 2 item attempts remaining 8:03Explanation / Answer
Answer:
The reaction is
Iron(II) chloride(aq)+Silver nitrate(aq) --------> Iron (II) nitrate(aq)+silver chloride(s).
FeCl2(aq)+AgNO3(aq)------->Fe(NO3)2(aq)+AgCl(s)
The balanced equation is
FeCl2(aq)+2AgNO3(aq) -------> Fe(NO3)2(aq)+2AgCl(s)
Given mass of silver nitrate=26.6 g and
Molar mass of AgNO3=169.87 g/mol.
Moles of AgNO3=mass/molar mass=26.6 g/169.87 g/mol
Moles of AgNO3=0.1566 mol.
The amount of FeCl2 is excess then the limiting reagent is AgNO3.
The mole ratio between AgNO3 to Fe(NO3)2 is 2:1
So the moles of Fe(NO3)2=1/2(mol AgNO3)=0.1566 mol/2=0.0783 mol.
Molar mass of Fe(NO3) 2=179.89 g/mol.
Mass of Fe(NO3)2=moles X molar mass
Mass of Fe(NO3)2=0.0783 mol X 179.89 g/mol=14.08 g.
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