Cisplatin [Pt(NH3)2Cl2], an anticancer agent used for the treatment of solid tumors, is prepared by the reaction of ammonia (NH3) with potassium tetrachloroplatinate (K2PtCl4). Assume that 10.0 g of K2PtCl4 and 10.0 g of NH3 are allowed to react. K2PtCl4 + 2 NH3 ï‚® Pt(NH3)2Cl2 + 2 KCl a) Which reactant is limiting, and which is in excess? b) How many grams of the excess reactant are consumed, and how many grams remain? c) How many grams of cisplatin are formed?
Cisplatin [Pt(NH3)2Cl2], an anticancer agent used for the treatment of
solid tumors, is prepared by the reaction of ammonia (NH3) with potassium
tetrachloroplatinate (K2PtCl4). Assume that 10.0 g of K2PtCl4 and 10.0 g of NH3 are
allowed to react.
K2PtCl4 + 2 NH3 ï‚® Pt(NH3)2Cl2 + 2 KCl
a) Which reactant is limiting, and which is in excess?
b) How many grams of the excess reactant are consumed, and how many grams remain?
c) How many grams of cisplatin are formed?
The Correct Answer and Explanation is:
Balanced Chemical Equation:
K2PtCl4+2NH3→Pt(NH3)2Cl2+2KCl\text{K}_2\text{PtCl}_4 + 2 \text{NH}_3 \rightarrow \text{Pt(NH}_3)_2\text{Cl}_2 + 2 \text{KCl}
Molar Masses:
- K₂PtCl₄ = 2(39.1) + 195.1 + 4(35.5) = 415.1 g/mol
- NH₃ = 14.0 + 3(1.0) = 17.0 g/mol
- Cisplatin [Pt(NH₃)₂Cl₂] = 195.1 + 2(17.0) + 2(35.5) = 300.1 g/mol
Moles of Each Reactant:
- K₂PtCl₄:
10.0 g415.1 g/mol≈0.0241 mol\frac{10.0 \text{ g}}{415.1 \text{ g/mol}} \approx 0.0241 \text{ mol} - NH₃:
10.0 g17.0 g/mol≈0.588 mol\frac{10.0 \text{ g}}{17.0 \text{ g/mol}} \approx 0.588 \text{ mol}
Limiting Reactant:
From the balanced equation:
- 1 mol of K₂PtCl₄ reacts with 2 mol of NH₃.
- For 0.0241 mol of K₂PtCl₄, we need:
0.0241×2=0.0482 mol NH30.0241 \times 2 = 0.0482 \text{ mol NH}_3
We have 0.588 mol NH₃, which is much more than needed.
✅ K₂PtCl₄ is the limiting reactant
✅ NH₃ is in excess
(a) Answer:
- Limiting reactant: K₂PtCl₄
- Excess reactant: NH₃
(b) Grams of NH₃ Consumed and Remaining:
- NH₃ consumed = 2 × 0.0241 mol = 0.0482 mol
- Grams consumed = 0.0482 mol × 17.0 g/mol = 0.82 g
- Grams remaining = 10.0 g – 0.82 g = 9.18 g
(c) Grams of Cisplatin Formed:
- From 1 mol K₂PtCl₄ → 1 mol cisplatin
- So, 0.0241 mol K₂PtCl₄ → 0.0241 mol cisplatin
- Mass = 0.0241 mol × 300.1 g/mol = 7.23 g
Final Answers:
a) Limiting reactant: K₂PtCl₄, Excess: NH₃
b) NH₃ consumed: 0.82 g, NH₃ remaining: 9.18 g
c) Mass of cisplatin formed: 7.
The preparation of cisplatin involves a simple stoichiometric reaction between potassium tetrachloroplatinate (K₂PtCl₄) and ammonia (NH₃). To determine the limiting reagent and calculate the product mass, we start by analyzing the chemical reaction: K2PtCl4+2NH3→Pt(NH3)2Cl2+2KCl\text{K}_2\text{PtCl}_4 + 2 \text{NH}_3 \rightarrow \text{Pt(NH}_3)_2\text{Cl}_2 + 2 \text{KCl}
This equation shows that 1 mole of K₂PtCl₄ reacts with 2 moles of NH₃ to produce 1 mole of cisplatin. The first step is to convert the given masses into moles using their molar masses. For K₂PtCl₄, we calculate about 0.0241 mol, and for NH₃, we have about 0.588 mol.
Using stoichiometry, we find that the amount of NH₃ needed for the reaction is 0.0482 mol (2 × 0.0241 mol). Since 0.588 mol is available, which is significantly more than needed, NH₃ is the excess reactant. This makes K₂PtCl₄ the limiting reactant, as it restricts the amount of product that can be formed.
Next, we determine the mass of NH₃ consumed: 0.0482 mol × 17.0 g/mol = 0.82 g. Therefore, 10.0 g – 0.82 g = 9.18 g NH₃ remains unreacted.
Finally, using the 1:1 molar ratio between K₂PtCl₄ and cisplatin, we find that 0.0241 mol of K₂PtCl₄ will produce 0.0241 mol of cisplatin. This equates to a mass of 7.23 g.
This stoichiometric analysis is crucial in synthetic chemistry and pharmaceutical manufacturing, ensuring efficient resource use and accurate prediction of product yield.
