Question

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?

Answer 1

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.