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The chemotherapy drug cis-platin, Pt(NH3)2Cl2, is prepared via the chemical reaction shown below.

(NH4)2PtCl4 + NH3 ---> Pt(NH3)2Cl2 + NH4Cl

What volume of 0.400 M NH3 (ammonia) is required to completely react with 25.00 mL of a 0.100 M solution of the platinum-containing starting compound, (NH4)2PtCl4?

User Jpoh
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1 Answer

3 votes

Answer:

12.5 mL NH₃

Step-by-step explanation:

To find the volume of NH₃, you need to (1) find the moles (NH₄)₂PtCl₄ (using the molarity equation), then (2) convert moles (NH₄)₂PtCl₄ to moles NH₃ (via mole-to-mole ratio from balanced equation, and then (3) find the volume NH₃ (using the molarity equation).

(Step 1)

25.00 mL / 1,000 = 0.025 L

Molarity = moles / volume (L)

0.100 M = moles / 0.025 L

0.0025 = moles (NH₄)₂PtCl₄

(Step 2)

The unbalanced equation:

(NH₄)₂PtCl₄ + NH₃ ----> Pt(NH₃)₂Cl₂ + NH₄Cl

Reactants: 2 nitrogen, 11 hydrogen, 1 platinum, 4 chlorine

Products: 2 nitrogen, 10 hydrogen, 1 platinum, 3 chlorine

The balanced equation:

(NH₄)₂PtCl₄ + 2 NH₃ ----> Pt(NH₃)₂Cl₂ + 2 NH₄Cl

Reactants: 3 nitrogen, 14 hydrogen, 1 platinum, 4 chlorine

Products: 2 nitrogen, 14 hydrogen, 1 platinum, 4 chlorine

0.0025 moles (NH₄)₂PtCl₄ 2 moles NH₃
---------------------------------------- x ------------------------------ = 0.0050 moles NH
1 mole (NH₄)₂PtCl₄

(Step 3)

Molarity = moles / volume (L)

0.400 M = 0.0050 moles / volume

(0.400 M) x volume = 0.0050 moles

volume = 0.0125 L NH

0.0125 L x 1,000 = 12.5 mL NH

User Chris Colla
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