Respuesta :
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.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(1)
[tex]\text{Number of moles of zinc}=\frac{3.00g}{65g/mol}=0.0462mol[/tex]
[tex]\text{Number of moles of silver nitrate}=\frac{3.75g}{170g/mol}=0.0220mol[/tex]
The chemical equation is:
[tex]Zn(s)+2AgNO_3(aq)\rightarrow Zn(NO_3)_2(aq)+1Ag(s)[/tex]
By stoichiometry of the reaction;
2 moles of silver nitrate reacts with 1 mole of zinc
Thus 0.0220 moles silver nitrate react with=[tex]\frac{1}{2}\times 0.0220=0.0110[/tex] 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=[tex]\frac{2}{2}\times 0.0220=0.0220[/tex] moles of silver
[tex]0.0220mol=\frac{\text{Mass of silver}}{108g/mol}\\\\\text{Mass of silver}=2.38g[/tex]
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.
