I think D im not completely sure
The concentration of AlCl3 solution if 150 ml of the solution contains 550 mg of cl- ion is 0.0344 M
calculation
concentration = moles /volume in liters
volume in liters = 150 /1000= 0.15 L
number of moles calculation
write the equation for dissociation of Al2Cl3
that is AlCl3 ⇔ Al^3+ + 3 Cl ^-
find the moles of Cl^- formed
moles =mass/molar mass
mass in grams= 550/ 1000 =0.55 grams
molar mass of Cl^- =35.5 g/mol
moles is therefore= 0.55/35.5 =0.0155 moles
by use of mole ration betweem AlCl3 to Cl^- which is 1:3 the moles of AlCl3 is =0.0155 x 1/3= 5.167 x10^-3 moles
concentration of AlCl3 is therefore= 5.167 x10^-3/ 0.15 =0.0344 M
Answer:
well if you google how to find predicting product you will find the answer i dont think this counts as my answer but i tryed
Explanation:
Answer:
14.33 g
Explanation:
Solve this problem based on the stoichiometry of the reaction.
To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and finally calculate the mass of AgCl.
2 AgNO₃ + CaCl₂ ⇒ Ca(NO₃)₂ + 2 AgCl
mass, g 6.97 6.39 ?
MW ,g/mol 169.87 110.98 143.32
mol =m/MW 0.10 0.06 0.10
From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :
0.10 mol x 143.32 g/mol = 14.33 g
