The number of liters of 3.00 M lead (II) iodide : 0.277 L
<h3>Further explanation</h3>Reaction(balanced)
Pb(NO₃)₂(aq) + 2KI(aq) → 2KNO₃(aq) + PbI₂(s)
moles of KI = 1.66
From the equation, mol ratio of KI : PbI₂ = 2 : 1, so mol PbI₂ :
Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution
Where
M = Molarity
n = Number of moles of solute
V = Volume of solution
So the number of liters(V) of 3.00 M lead (II) iodide-PbI₂ (n=0.83, M=3):
Answer:
93,32 % (w/w)
Explanation:
The Ca²⁺ of CaCO₃ was completely precipitate to CaC₂O₄ that results in H₂C₂O₄. The titration of this one is:
3H₂C₂O₄ + 2H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O + 6CO₂
As you required 35,62mL of 0,1092M MnO₄⁻, the moles of H₂C₂O₄ are:
0,03562L× =
3,890x10⁻³mol of MnO₄⁻× = <em>0,01167 mol of H₂C₂O₄</em>
The number of moles of CaCO₃ are the same number of moles of H₂C₂O₄ because every Ca²⁺ was converted in CaC₂O₄ that was converted in H₂C₂O₄. That means: <em>0,01167 mol of CaCO₃</em>
0,01167 mol of CaCO₃ are:
0,01167 mol of CaCO₃× = <em>1,168 g of CaCO₃</em>
As the mass of the initial mixture is 1,2516 g, the percentage by weight of CaCO₃ is:
×100 = <em>93,32 % (w/w)</em>
I hope it helps!