Cisplatin, PtCl2(NH3)2, ?is a drug used in the treatment of cancer. It can be synthesized by the reaction of K2PtCl4 ?with ammonia. What mass of ammonia in grams is needed to produce 1.00kg ?of cisplatin?
The Correct Answer and Explanation is:
To determine the mass of ammonia (NH₃) needed to produce 1.00 kg (1000 g) of cisplatin (PtCl₂(NH₃)₂), we must first use stoichiometry based on the balanced chemical reaction and molar masses.
Step 1: Balanced Chemical Equation
The synthesis of cisplatin from potassium tetrachloroplatinate(II) and ammonia is: K2PtCl4+2NH3→PtCl2(NH3)2+2KCl\text{K}_2\text{PtCl}_4 + 2\text{NH}_3 \rightarrow \text{PtCl}_2(\text{NH}_3)_2 + 2\text{KCl}
From the equation, 2 moles of NH₃ are required to form 1 mole of cisplatin.
Step 2: Molar Masses
- PtCl₂(NH₃)₂ (cisplatin):
- Pt = 195.08 g/mol
- Cl₂ = 2 × 35.45 = 70.90 g/mol
- NH₃ = 2 × (14.01 + 3 × 1.008) = 2 × 17.034 = 34.068 g/mol
- Total = 195.08 + 70.90 + 34.068 = 300.05 g/mol
- NH₃:
- 14.01 + (3 × 1.008) = 17.034 g/mol
Step 3: Calculate Moles of Cisplatin
Moles of cisplatin=1000 g300.05 g/mol≈3.33 mol\text{Moles of cisplatin} = \frac{1000\, \text{g}}{300.05\, \text{g/mol}} \approx 3.33\, \text{mol}
Step 4: Use Mole Ratio to Get Moles of NH₃
From the balanced equation: 1 mol cisplatin⇒2 mol NH₃1\, \text{mol cisplatin} \Rightarrow 2\, \text{mol NH₃} 3.33 mol cisplatin⇒2×3.33=6.66 mol NH₃3.33\, \text{mol cisplatin} \Rightarrow 2 \times 3.33 = 6.66\, \text{mol NH₃}
Step 5: Convert Moles of NH₃ to Grams
Mass NH₃=6.66 mol×17.034 g/mol≈113.5 g\text{Mass NH₃} = 6.66\, \text{mol} \times 17.034\, \text{g/mol} \approx 113.5\, \text{g}
✅ Final Answer: 113.5 grams of ammonia are needed to produce 1.00 kg of cisplatin.
Explanation
Cisplatin, PtCl₂(NH₃)₂, is a platinum-based chemotherapy drug that works by binding to DNA in cancer cells, disrupting their function and leading to cell death. To synthesize cisplatin in the lab or industrially, potassium tetrachloroplatinate(II) (K₂PtCl₄) is reacted with ammonia (NH₃). The stoichiometry of the reaction reveals a 1:2 molar ratio between cisplatin and ammonia: one mole of cisplatin forms from two moles of ammonia.
To determine how much ammonia is required to produce 1.00 kg (1000 g) of cisplatin, we first calculate the number of moles of cisplatin needed. Using its molar mass (300.05 g/mol), we find that 1000 g corresponds to approximately 3.33 moles of cisplatin. Given the 2:1 ratio, we need 6.66 moles of ammonia.
Next, we convert these moles into grams using the molar mass of ammonia, which is approximately 17.034 g/mol. Multiplying 6.66 moles by this value yields roughly 113.5 grams of ammonia.
This calculation is essential in synthetic chemistry and pharmaceutical manufacturing to ensure efficient resource use and avoid excess reactants. The precise measurement of reactants also minimizes waste and cost while optimizing product yield and purity—crucial factors in the production of high-quality drugs like cisplatin.
