Answer to Question #158177 in General Chemistry for Maria

Question

BaCl₂ + (NH₄)₂CO₃ --> BaCO₃ + NH₄Cl

120ml of 0.6M BaCl is added to 650ml of 0.45M (NH₄)₂CO₃. How many grams of BaCO₃ is made? What is the concentration of left over reactant in the water?

Answer 1

Molar mass of BaCl2= 208.34g/mol

Molar mass of (NH4)2CO3= 96g/mol

Molar mass of BaCO3= 197.34g/mol

Mole of BaCl2= C xV

= 0.6 x 120/1000 = 0.072mol

Mass of BaCl2= mole x molar mass

= 0.072 x 208.34

= 15.0g

Mole of (NH4)2CO3= C xV

= 0.45 x 650/1000

= 0.29mole

Mass of (NH4)2CO3= 0.29 x 96

=27.84

Now to know the mass of BaCO3, let us find the limiting reagent.

96g of (NH4)2CO3 reacts with 208.34g of BaCl2

27.84g of (NH4)2CO3 should react with 208.34 x 27.84/96 = 60.42g

Since only 15.0g of BaCl2 is available, it is the limiting reagent.

Now,

208.34g of BaCl2 gives 197.34g of BaCO3

15.0g of BaCl2 will give 197.34 x 15.0/208.34 = 14.21g of BaCO3

To find the left over reactant

208.34g of BaCl2 reacts with 96g of (NH3)2CO3

15.0g of BaCl2 will react with 96 x 15.0/208.34

= 6.91

Mass of left over reactant = 27.84-6.91 = 20.9g

Total volume= 120ml + 650ml

= 770ml= 0.77L

Mass concentration= 20.9/0.77= 27.14gdm^-3

Molar concentration= mass concentration/molar mass = 27.14/96 = 0.28M