Consider the reaction: Cu (s) + 2 AgNO3 (aq) = 2 Ag (s) + Cu (NO3) 2 (aq)

If 3.00 moles of copper metal react:

a) How many moles of silver metal will be produced?

b) How many grams of silver metal will be produced?

c) How many moles of copper (II) nitrate will be produced?

d) How many grams of copper (II) nitrate will be produced?

a)

According to the reaction equation:

1 mole of copper produces two moles of silver metal

3 moles of copper metal will produce 3*2/1 = 6 moles of silver metal

b)

Mass of silver produced = number of moles * molar mass

Mass of silver produced = 6*108 = 648g of silver

c)

1 mole of Cu produced 1 mole of copper II nitrate

3 moles of copper will produce 3*1/1 = 3moles of copper Ii nitrate

d)

Molar mass of copper II nitrate = 64 + 2 (14+3 (16)) = 188gmol-1

Mass of copper II nitrate = 3 * 188 = 564g of copper II nitrate

(a) 6 moles of Ag

(b) 647.22g of Ag

(c) 3 moles of copper (II) nitrate

(d) 564g of copper (II) nitrate

Explanation:

Cu (s) + 2AgNO3 (aq) = 2Ag (s) + Cu (NO3) 2 (aq)

(a) According to the above equation, 1 mole of Cu gives 2 mole of Ag. So, 3 moles of Cu will give:

Cu : Ag

1 2

3 x

x = 3*2

x = 6 moles of silver

(b) To find grams of silver metal, multiply the moles from (a) with its molar mass,

mass=no. of moles * MM

= 6 * 107.87g

= 647.22g of Ag will be produced

(c) According to the above equation, 1 mole of Cu gives 1 mole of Cu (NO3) 2, so 3 moles of Cu will give:

Cu : Cu (NO3) 2

1 1

3 x

x = 3 * 1

x = 3 moles of Cu (NO3) 2

(d) To find grams of Cu (NO3) 2, multiple moles with (a) with molar mass of Cu (NO3) 2,

mass of Cu (NO3) 2 = moles * MM

= 3 * 188g/mol

= 564g of Cu (NO3) 2