Answer:
Zn + 2AgNO₃ → Zn(NO₃)₂ +2Ag
13.34g Zn(NO₃)₃ and 15.24g Ag are formed if reaction is 100% complete.
Explanation:
molar mass of Zn=65 g
molar mass of AgNO₃=170 g
molar mass of Zn(NO₃)₂ =189 g
molar mass of Silver = 108 g
Zn + 2 AgNO₃ → Zn(NO₃)₂ +2 Ag eq(1)
1 mole Zn reacts with 2 moles Silver nitrate to give 1 mole Zinc nitrate and 2 moles Silver
or
65 g Zn reacts with 2×170 g of AgNO₃ → 189 g Zn(NO₃)₂ and 2×108 g Ag - eq(2)
First find limiting reagent of the reaction
65g Zn reacts with 2×170g of AgNO₃
12 g Zn reacts with (12÷65)×2×170 g AgNO₃
=62.7 g AgNO₃
For the reaction to go to 100% yield 12 g Zn will need 62.7 g AgNO₃
but amount of AgNO₃ is 24 g
So the reaction yields is limited by amount of AgNO₃.
AgNO₃ is the limiting reagent.
So calculate the yield of products with the amount of AgNO₃
by eq(2)
2× 170 gAgNO₃ gives 189 g Zn(NO₃)₃
=340 g AgNO₃ gives 189 g Zn(NO₃)₃
24 g AgNO₃ gives (24÷340) ×189 g Zn(NO₃)₃
=13.34g Zn(NO₃)₃
again by eq2
2×170 g AgNO₃ gives 2×108 g Ag
= 340 g AgNO₃ gives 216 g Ag
24 g AgNO₃ gives (24÷340)×216 g Ag
= 15.24g Ag
