What volume (L) of 0.250 M KCl will completely react with 0.170 L of a 0.235...

What volume (L) of 0.250 M KCl will completely react with 0.170 L of a 0.235 M solution of Pb(NO3)2 according to the balanced equation shown below? 2 KCl (aq) + Pb(NO3)2 (aq) = PbCl2 (s) + 2 KNO3 (aq)

Expert Solution

The balanced equation

2 KCl (aq) + Pb(NO₃)₂ (aq) = PbCl₂ (s) + 2 KNO₃ (aq)

Step 1- Convert lead nitrate to moles

Moles of lead nitrate = Concentration of lead nitrate x Vol of lead nitrate in L

= 0.235 x 0.170 = 0.0399 moles

Step 2: Use the mole ratio between potassium chloride and lead nitrate to calculate how many moles of KCl is needed

According to reaction stoichiometry for every mole of lead nitrate, 2 moles of KCl is needed

So, for 0.02625 moles of lead nitrate twice of moles of KCl is needed

Thus, the moles of KCl =2* 0.0399 moles = 0.0798 moles

Step3- Find the volume of KCl –

V =n/C

V = 0.0798 moles/0.250 M

V = 0.3192L

V = 319.2 ml

queen_honey_blossom answered 2 years ago

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