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 K2PtCl4 and 10.0 g of NH3 are allowed to react.

Question

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 [latex]K_{2}PtCl_{4}[/latex] and 10.0 g of [latex]NH_{3}[/latex] are allowed to react.

[latex]\underset{Potassium tetrachloroplatinate}{K_{2}PtCl_{4}(aq)} + 2 NH_{3}(aq) → \underset{Cisplatin}{Pt(NH_{3})_{2}Cl_{2}(s)} + 2 KCl(aq)[/latex]

(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).

Accepted Answer

(a) Finding the molar amounts of reactants always begins by calculating formula masses and using molar masses as conversion factors:

Form. mass of [latex]K_{2}PtCl_{4}[/latex] = (2 × 39.1 amu) + 195.1 amu + (4 × 35.5 amu) = 415.3 amu

Molar mass of [latex]K_{2}PtCl_{4}[/latex] = 415.3 g/mol

Moles of [latex]K_{2}PtCl_{4}[/latex] = [latex]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}[/latex]

Molec. mass of [latex]NH_{3}[/latex] = 14.0 amu + (3 × 1.0 amu) = 17.0 amu

Molar mass of [latex]NH_{3}[/latex] = 17.0 g/mol

Moles of [latex]NH_{3}[/latex] = [latex]10.0 \cancel{g NH_{3}} × \frac{1 mol NH_{3}}{17.0 \cancel{g NH_{3}}} = 0.588 mol NH_{3}[/latex]

These calculations tell us that we have 0.588 mol of ammonia and 0.0241 mol of [latex]K_{2}PtCl_{4}[/latex], or 0.588/0.0241 = 24.4 times as much ammonia as [latex]K_{2}PtCl_{4}[/latex]. The coefficients in the balanced equation, however, say that only two times as much ammonia as [latex]K_{2}PtCl_{4}[/latex] is needed. Thus, a large excess of [latex]NH_{3}[/latex] is present, and [latex]K_{2}PtCl_{4}[/latex] 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 ([latex]K_{2}PtCl_{4}[/latex]) is used up, but only the amount of the excess reactant ( [latex]NH_{3}[/latex]) required by stoichiometry undergoes reaction:

Moles of [latex]K_{2}PtCl_{4}[/latex] consumed = 0.0241 mol [latex]K_{2}PtCl_{4}[/latex]

[latex]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}[/latex]

Grams of N[latex]H_{3} consumed = 0.0482 \cancel{mol NH_{3}} × \frac{17.0 g NH_{3}}{1 \cancel{mol NH_{3}}} = 0.819 g NH_{3}[/latex]

Grams of [latex]NH_{3}[/latex] not consumed = (10.0 g - 0.819 g)[latex]NH_{3}[/latex] = 9.2 g [latex]NH_{3}[/latex]

(c) The balanced equation shows that 1 mol of cisplatin is formed for each mole of [latex]K_{2}PtCl_{4}[/latex] consumed. Thus, 0.0241 mol of cisplatin is formed from 0.0241 mol of [latex]K_{2}PtCl_{4}[/latex]. 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 [latex]Pt(NH_{3})_{2}Cl_{2}[/latex] = 195.1 amu + (2 × 17.0 amu) + (2 × 35.5 amu) = 300.1 amu

Molar mass of [latex]Pt(NH_{3})_{2}Cl_{2}[/latex] = 300.1 g/mol

[latex]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}[/latex]

Question 3.10: CALCULATING THE AMOUNT OF AN EXCESS REACTANT Cisplatin, an a...

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