K2PtCl4 + 2NH3 → 2KCl + PtCl2(NH3)2 Determine the mass of cisplatin produced from 2.5 g K2PtCl4 reacting with excess ammonia

Question

Some types of cancer can be effectively treated by a compound called cisplatin, PtCl₂(NH₃)₂. Cisplatin is produced synthetically by the following balanced equation:

K₂PtCl₄ + 2NH₃ → 2KCl + PtCl₂(NH₃)₂

Determine the mass of cisplatin produced from 2.5 g K₂PtCl₄ reacting with excess ammonia (excess ammonia means that you have plenty of ammonia to react with cisplatin).

Answer 1

Savanna: I see that you have posted about 8 or more questions of this same type and I'll be happy to answer 1 or 2 but I doubt you'll get all of them answered. You should learn from the ones that are done, and do the others yourself.

ALWAYS start with a correctly balanced equation:

K₂PtCl₄ + 2NH₃ → 2KCl + PtCl₂(NH₃)₂ .. balanced equation

Since there is excess NH3, we don't have to be concerned with limiting reactants. We simply use the moles of K₂PtCl₄ to find the moles of PtCl₂(NH₃)₂ and then use the molar mass of PtCl₂(NH₃)₂ to convert to grams.

Molar mass K₂PtCl₄ = 415 g / mole

Molar mass PtCl₂(NH₃)₂ = 300 g / mole

moles K₂PtCl₄ present = 2.5 g x 1 mol / 415 g = 0.00602 moles

moles PtCl₂(NH₃)₂ produced = 0.00602 mols K₂PtCl₄ x 1 mol PtCl₂(NH₃)₂ / mol K₂PtCl₄ = 0.00602 mols

mass cisplatin = 0.00602 mols cisplatin x 300 g / mole = 1.81 g cisplatin produced