If 5.85 grams of cobalt metal react with 15.8 grams of silver nitrate, how many grams of silver metal can be formed and how many grams of the excess reactant will be left over when the reaction is complete? show all of your work. unbalanced equation: co + agno3 “yields”/ co(no3)2 + ag

Answers:
Answer 1: 10.03 g of siver metal can be formed.
Answer 2: 3.11 g of Co are left over.

Work:

1) Unbalanced chemical equation (given):

Co + AgNO3 → Co(NO3)2 + Ag

2) Balanced chemical equation

Co + 2AgNO3 → Co(NO3)2 + 2Ag

3) mole ratios

1 mol Co : 2 mole AgNO3 : 1 mol Co(NO3)2 : 2 mol Ag

4) Convert the masses in grams of the reactants into number of moles

4.1) 5.85 grams of Co

# moles = mass in grams / atomic mass

atomic mass of Co = 58.933 g/mol

# moles Co = 5.85 g / 58.933 g/mol = 0.0993 mol

4.2) 15.8 grams of Ag(NO3)

# moles Ag(NO3) = mass in grams / molar mass

molar mass AgNO3 = 169.87 g/mol

# moles Ag(NO3) = 15.8 g / 169.87 g/mol = 0.0930 mol

5) Limiting reactant

Given the mole ratio 1 mol Co : 2 mol Ag(NO3) you can conclude that there is not enough Ag(NO3) to make all the Co react.

That means that Ag(NO3) is the limiting reactant, which means that it will be consumed completely, whilce Co is the excess reactant.

6) Product formed.

Use this proportion:

2 mol Ag(NO3)           0.0930mol Ag(NO3)    
=     
    2 mol Ag                              x

=> x = 0.0930 mol

Convert 0.0930 mol Ag to grams:

mass Ag = # moles * atomic mass = 0.0930 mol * 107.868 g/mol = 10.03 g

Answer 1: 10.03 g of siver metal can be formed.

6) Excess reactant left over

    1 mol Co                             x
= 
2 mole Ag(NO3)       0.0930 mol Ag(NO3)

=> x = 0.0930 / 2 mol Co = 0.0465 mol Co reacted

Excess = 0.0993 mol - 0.0465 mol = 0.0528 mol

Convert to grams:

0.0528 mol * 58.933 g/mol = 3.11 g

Answer 2: 3.11 g of Co are left over.