Example: A 2.00 gram piece of zinc is placed in 2.50 grams of aqueous silver nitrate. a) Write the balanced equation. b) Determine the limiting reagent. c) H

Question

Example: A 2.00 gram piece of zinc is placed in 2.50 grams of aqueous silver nitrate.
a) Write the balanced equation.
b) Determine the limiting reagent.
c) How many grams of silver will be formed?
d) How many grams of the excess reactant will be left over?

Answer 1

(a)  Zn(s) + 2AgNO₃(aq) ==> Zn(NO₃)₂(aq) + 2Ag(s)

(b)  Assuming that the problem means 2.5 g AgNO₃ (not 2.5 g of solution), then ...

moles AgNO₃ = 2.5 g AgNO₃ x 1 mole/169.87 g = 0.0147 moles AgNO₃

moles Zn = 2.00 g x 1 mole/65.38 g = 0.0306 moles Zn

Limiting reactant = AgNO₃ (see mole ratio of Zn:AgNO₃)

(c)  Since AgNO₃ is limiting, and 2 moles AgNO₃ = 1 mole Ag, 0.0147 moles AgNO₃ = 0.00735 moles Ag. Convert to mass using atomic mass of Ag.

(d)  Zn is in excess.  0.0147 moles AgNO₃ will use up 1/2 that many moles of Zn (see balanced reaction mole ratios). Convert that # moles to grams and subtract that amount from the initial amount of Zn (2.00 g).  This will give you grams of Zn left over.,