Question

consider the reaction between strontium chloride and silver nitrate. Write the overall reaction that occurs as well as the net ionic equation. calculate the number of grams of silver chloride produced from the complete reaction of 50.0 mL of .18 M strontium chloride.

4. Practice: Consider the reaction between strontium chloride and silver (1) nitrate. Write the overall reaction that occurs as well as the net ionic reaction. Calculate the number of grams of silver (1) chloride produced from complete reaction of 50.0 ml. of 0.18 M strontium chloride.

Answer 1

4.Ans :

Overall reaction between SrCl2 and AgNO3 is : SrCl2 (aq) + 2 AgNO3 (aq) ---------------------> 2 AgCl (s) + Sr(NO3)2 (aq)

Complete ionic equation between SrCl₂ and AgNO₃ is :

Sr²⁺ (aq) + 2 Cl^- (aq) + 2 Ag⁺ (aq) + 2 NO₃^- (aq) ---------> 2 AgCl (s) + Sr²⁺ (aq) + 2 NO₃^- (aq)

Net ionic equation can be obtain by cancel out the spectator ions (Sr2+ and NO3-) on both the side of the equation : 2 Ag+ (aq) + 2 Cl- (aq) --------> 2 AgCl (s)

From Molarity definition :

Molarity = Number of moles / Volume in L

Therefore,

Number of moles of SrCl₂ = Molarity of SrCl₂ x Volume of SrCl₂ in L = 0.18 M x 0.050 L = 0.009 mol

From the balance chemical equation :

1 mole of SrCl₂ gives = 2 mol of AgCl (s)

and

0.009 mol of SrCl₂ gives = 2 mol x 0.009 mol / 1 mol = 0.018 mol of AgCl (s)

Number of moles = Given mass in g / Gram molar mass

Therefore,

Mass of AgCl (s) formed = Moles x Gram molar mass of AgCl = 0.018 mol x 143.32 g/mol = 2.58 g

Therefore,

Mass of AgCl (s) in gram formed in the reaction = 2.58 g