Mass of cisplatin can be made : 144.6 g
<h3>Further explanation</h3>
Given
Reaction(unbalanced)
K2PtCl4 (aq) + ___NH3 (aq) ___Pt(NH3)2Cl2 (s) + ___KCl (aq)
200 g of K2PtCl4
Required
mass of cisplatin Pt(NH3)2Cl2
Solution
Balanced equation
K₂PtCl₄ (aq) + 2NH₃ (aq) ⇒Pt(NH₃)₂Cl₂ (s) + 2KCl (aq)
mol of K₂PtCl₄(MW=2.39+195+4.35.5=415 g/mol) :
mol = mass : Mw
mol = 200 : 415
mol = 0.482
From equation, mol ratio K₂PtCl₄ : Pt(NH₃)₂Cl₂ = 1 : 1, so mol Pt(NH₃)₂Cl₂ = 0.482
Mass of cisplatin Pt(NH₃)₂Cl₂(MW=195+2.14+6.1+2.35.5=300 g/mol) :
mass = mol x MW
mass = 0.482 x 300
mass = 144.6 g
Answer:
0.00125 moles H₃X
Solution and Explanation:
In this question we are required to calculate the number of moles of triprotic acid neutralized in the titration.
Volume of NaOH used = final burette reading - initial burette reading
= 39.18 ml - 3.19 ml
= 35.99 ml or 0.03599 L
Step 1: Moles of NaOH used
Number of moles = Molarity × Volume
Molarity of NaOH = 0.1041 M
Moles of NaOH = 0.1041 M × 0.03599 L
= 0.00375 mole
Step 2: Balanced equation for the reaction between triprotic acid and NaOH
The balanced equation is;
H₃X(aq) + 3NaOH(aq) → Na₃X(aq) + 3H₂O(l)
Step 3: Moles of the triprotic acid (H₃X used
From the balanced equation;
1 mole of the triprotic acid reacts with 3 moles of NaOH
Therefore; the mole ratio of H₃X to NaOH is 1 : 3.
Therefore;
Moles of Triprotic acid = 0.00375 mole ÷ 3
= 0.00125 moles
Hence, moles of triprotic acid neutralized during the titration is 0.00125 moles.
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Answer:
A
Explanation:
I just did this question in class:))