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# How many grams of silver can be produced from the reaction of 92.8 g of silver nitrate with 1.34 g of aluminum?

Updated: 4/28/2022

Wiki User

6y ago

Oxidation-reduction reaction:Ag^+(aq) + Al(s) ===> Ag(s) + Al^3+ or looked at another way...

3AgNO3(aq) + Al(s) ===> Al(NO3)3(aq) + 3Ag(s)

moles AgNO3 present = 92.8 g x 1 mole/170 g =0.546 moles

moles Al present = 1.34 g x 1 mole/26.9 g = 0.0498 moles

Al is limiting based on mole ratio of 3 AgNO3 : 1 Al

moles Ag(s) produced = 0.0498 moles Al x 3 moles Ag/mole Al = 0.1494 moles Ag

mass of Ag = 0.1494 moles Ag x 108 g/mole = 16.1 g Ag formed

Wiki User

6y ago

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Q: How many grams of silver can be produced from the reaction of 92.8 g of silver nitrate with 1.34 g of aluminum?
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