A mixture consisting of only z
A mixture consisting of only zinc(II) chloride (ZnCl2, 136.31g/mol) and titanium(IV) chloride (TiCl4, 189.67 g/mol) weighs1.0971 g. When the mixture is dissolved in water and an excess ofsilver acetate is added, all the chloride ions associated with theoriginal mixture are precipitated as insoluble silver chloride(AgCl, 143.32 g/mol). The mass of the silver chloride is found tobe 2.9691 g. Calculate the mass percentage of zinc(II) chloride inthe original mixture.
Answer:
Suppose the mixture contains x grams of ZnCl2 and ygrams of TiCl4 .
SO, x + y = 1.0971……………………………………………………………Equation1
Moles of ZnCl2 present : mass / molar mass ofZnCl2 = x g/ (136.31 g/mol) = 0.00734x mole
The reaction between ZnCl2 and silver acetate isgiven below:
ZnCl2 (aq) + 2 AgOCOCH3 (aq) =Zn(OCOCH3)2 (aq) + 2 AgCl (s)
From the balanced stoichiometry, we get:
Moles of AgCl : Moles of ZnCl2 = 2:1
So, moles of AgCl = 2 * moles of ZnCl2
So, in the reaction moles of AgCl produced = 2* 0.00734x moles =0.01468x mole
Moles of TiCl4 present : mass / molar mass ofTiCl4 = y g/ (189.67 g/mol) = 0.00527y mole
Similarly the reaction between TiCl4 and silveracetate is:
TiCl4 (aq) + 4 AgOCOCH3 (aq) =Ti(OCOCH3)4 (aq) + 4 AgCl (s)
From the balanced stoichiometry, we get:
Moles of AgCl : Moles of TiCl4 = 4:1
So, moles of AgCl = 4 * moles of TiCl4
So, in the reaction moles of AgCl produced = 4* 0.00527y moles =0.02108y mole
Total moles of AgCl produced = 0.01468x mole + 0.02108y mole
Mass of AgCl produced = Moles * molar mass of AgCl = (0.01468x +0.02108y) * 143.32 g
According to the question, mass of AgCl produced = 2.9691 g
So, (0.01468x + 0.02108y) * 143.32 g = 2.9691
Or, 0.01468x + 0.02108y = 0.0207
Substituting, y = 1.0971 -x from Equation 1 in the aboveequation, we get:
0.01468x + 0.02108* ( 1.0971 -x ) = 0.0207
Or, – 0.0064 x = – 0.0024
So, x = 0.375
So, in the mixture mass of ZnCl2 is 0.375 g
The mass percentage of ZnCl2 is : (0.375/ 1.0971) * 100% =34.18%
"Our Prices Start at $11.99. As Our First Client, Use Coupon Code GET15 to claim 15% Discount This Month!!"
