Cisplatin (Pt (NH3)2Cl2) is an anti-cancer drug used for the treatment of solid

Question

Cisplatin (Pt (NH3)2Cl2) is an anti-cancer drug used for the treatment of solid tumors. Cisplatin and potassium chloride (KC1) are produced when ammonia (NH3) is reacted with potassium tetrachloroplatinate (K2PtCl4) (ktcp) in the following reaction: K2PtCl4 (s) + NH3(g) (Pt (NH3)2C12)(S) + KCl(aq) Potassium Tetrachloroplatinate ammonia cisplatin potassium chloride ( ktep) FOR MOLAR MASSES USE TWO DECIMAL PLACES FOR ELEMENT MASSES In one process, 8.302 xlO5 nig of NH3 is reacted with 5.04 kg of ktep. Assume all the Limiting Reactant is converted cisplatin. Submit the following: The Balanced molecular equation, is : (Show the mole ratios here) Determine which reactant is the limiting reagent, Calculate the "theoretical "mass of cisplatin (g) formed (4.0 points) and; Calculate the excess reagent (in grams) that remains at the end of the reaction; In one reaction 3,543.0 grams of Cisplatin is produced . Calculate the percent yield.

Explanation / Answer

(1) Balanced molecular equation is:

K2PtCl4(s) + 2 NH3(g) => (Pt(NH3)2Cl2)(s) + 2 KCl(aq)

(2) Theoretical moles of ktcp : NH3 = 1 : 2

Moles of ktcp = mass/molar mass = 5.04 x 1000/415.09 = 12.142 mol

Moles of NH3 = mass/molar mass = 8.302 x 10^5 x 10^(-3)/17.03 = 47.164 mol

Actual moles of ktcp : NH3 = 12.142 : 47.164 = 1 : 3.884

Since NH3 is in excess => ktcp is the limiting reagent

(3) Moles of cisplatin = moles of ktcp = 12.142 mol

Theoretical mass of cisplatin = moles x molar mass of cisplatin

= 12.142 x 300.05

= 3643.2 g = 3643 g

(4) Moles of NH3 reacted = 2 x moles of ktcp

= 2 x 12.142 = 24.284 mol

Mass of NH3 reacted = moles x molar mass of NH3

= 24.284 x 17.03 = 413.6 g

Mass of NH3 remaining = 8.302 x 10^5 x 10^(-3) - 413.6

= 416.6 g = 417 g

(5) Percent yield = actual yield/theoretical yield x 100%

= 3543/3643 x 100%

= 97.25%

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