Question

(40 Points) Complete, balance, compute the amounts of the products assuming 100% yield. 12g Zinc Solid + 24g Silver Nitrate

Answer 1

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