Question 3.10: CALCULATING THE AMOUNT OF AN EXCESS REACTANT Cisplatin, an a...
CALCULATING THE AMOUNT OF AN EXCESS REACTANT
Cisplatin, an anticancer agent used for the treatment of solid tumors, is prepared by the reaction of ammonia with potassium tetrachloroplatinate. Assume that 10.0 g of K_{2}PtCl_{4} and 10.0 g of NH_{3} are allowed to react.
\underset{Potassium tetrachloroplatinate}{K_{2}PtCl_{4}(aq)} + 2 NH_{3}(aq) → \underset{Cisplatin}{Pt(NH_{3})_{2}Cl_{2}(s)} + 2 KCl(aq)(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?
STRATEGY
When solving a problem that deals with limiting reactants, the idea is to find how many moles of all reactants are actually present and compare the mole ratios of those actual amounts to the mole ratios required by the balanced equation. That comparison will identify the reactant there is too much of (the excess reactant) and the reactant there is too little of (the limiting reactant).
(a) Finding the molar amounts of reactants always begins by calculating formula masses and using molar masses as conversion factors:
Form. mass of K_{2}PtCl_{4} = (2 × 39.1 amu) + 195.1 amu + (4 × 35.5 amu) = 415.3 amu
Molar mass of K_{2}PtCl_{4} = 415.3 g/mol
Moles of K_{2}PtCl_{4} = 10.0 \cancel{g K_{2}PtCl_{4}} × \frac{1 mol K_{2}PtCl_{4}}{415.3 \cancel{g K_{2}PtCl_{4}}} = 0.0241 mol K_{2}PtCl_{4}
Molec. mass of NH_{3} = 14.0 amu + (3 × 1.0 amu) = 17.0 amu
Molar mass of NH_{3} = 17.0 g/mol
Moles of NH_{3} = 10.0 \cancel{g NH_{3}} × \frac{1 mol NH_{3}}{17.0 \cancel{g NH_{3}}} = 0.588 mol NH_{3}
These calculations tell us that we have 0.588 mol of ammonia and 0.0241 mol of K_{2}PtCl_{4}, or 0.588/0.0241 = 24.4 times as much ammonia as K_{2}PtCl_{4}. The coefficients in the balanced equation, however, say that only two times as much ammonia as K_{2}PtCl_{4} is needed. Thus, a large excess of NH_{3} is present, and K_{2}PtCl_{4} is the limiting reactant.
(b) With the identities of the excess reactant and limiting reactant known, we now have to find how many moles of each undergo reaction and then carry out mole-to-gram conversions to find the mass of each reactant consumed. The entire amount of the limiting reactant (K_{2}PtCl_{4}) is used up, but only the amount of the excess reactant ( NH_{3}) required by stoichiometry undergoes reaction:
Moles of K_{2}PtCl_{4} consumed = 0.0241 mol K_{2}PtCl_{4}
Moles of NH_{3} consumed = 0.0241 \cancel{mol K_{2}PtCl_{4}} × \frac{2 mol NH_{3}}{1 \cancel{mol K_{2}PtCl_{4}}} = 0.0482 mol NH_{3}
Grams of NH_{3} consumed = 0.0482 \cancel{mol NH_{3}} × \frac{17.0 g NH_{3}}{1 \cancel{mol NH_{3}}} = 0.819 g NH_{3}
Grams of NH_{3} not consumed = (10.0 g – 0.819 g)NH_{3} = 9.2 g NH_{3}
(c) The balanced equation shows that 1 mol of cisplatin is formed for each mole of K_{2}PtCl_{4} consumed. Thus, 0.0241 mol of cisplatin is formed from 0.0241 mol of K_{2}PtCl_{4}. To determine the mass of cisplatin produced, we must calculate its molar mass and then carry out a mole-to-gram conversion:
Molec. mass of Pt(NH_{3})_{2}Cl_{2} = 195.1 amu + (2 × 17.0 amu) + (2 × 35.5 amu) = 300.1 amu
Molar mass of Pt(NH_{3})_{2}Cl_{2} = 300.1 g/mol
Grams of Pt(NH_{3})_{2}Cl_{2} = 0.0241 \cancel{mol Pt(NH_{3})_{2}Cl_{2}} × \frac{300.1 g Pt(NH_{3})_{2}Cl_{2}}{1 \cancel{mol Pt(NH_{3})_{2}Cl_{2}}} = 7.23 g Pt(NH_{3})_{2}Cl_{2}