An amplifying device that indicates the presence of ionizing particles
1 answer:
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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!