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Zinc metal reacts with silver nitrate according to the reaction:

Zn(s) + 2AgNO3(aq)Zn(NO3 )2 (aq) + 2Ag(s)

Calculate the mass of Ag that forms when 3.00g of zinc metal is placed in an aqueous solution containing 3.75g of silver nitrate?

User EEP
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1 Answer

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Answer: 2.38 grams of silver forms when 3.00g of zinc metal is placed in an aqueous solution containing 3.75g of silver nitrate.

Step-by-step explanation:

To calculate the number of moles, we use the equation:


\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ....(1)


\text{Number of moles of zinc}=(3.00g)/(65g/mol)=0.0462mol


\text{Number of moles of silver nitrate}=(3.75g)/(170g/mol)=0.0220mol

The chemical equation is:


Zn(s)+2AgNO_3(aq)\rightarrow Zn(NO_3)_2(aq)+1Ag(s)

By stoichiometry of the reaction;

2 moles of silver nitrate reacts with 1 mole of zinc

Thus 0.0220 moles silver nitrate react with=
(1)/(2)* 0.0220=0.0110 moles of zinc

Thus silver nitrate will acts as limiting reagent and zinc acts as excess reagent.

2 moles of silver nitrate produces 2 mole of silver

Thus 0.0220 moles silver nitrate react with=
(2)/(2)* 0.0220=0.0220 moles of silver


0.0220mol=\frac{\text{Mass of silver}}{108g/mol}\\\\\text{Mass of silver}=2.38g

Thus 2.38 grams of silver forms when 3.00g of zinc metal is placed in an aqueous solution containing 3.75g of silver nitrate.

User Jimski
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