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Cisplatin $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$ is a cancer chemotherapy agent. Notice that it contains $\mathrm{NH}_{3}$ groups attached to platinum.

(a) What is the weight percent of $\mathrm{Pt}, \mathrm{N},$ and $\mathrm{Cl}$ in the cisplatin?

(b) Cisplatin is made by reacting $\mathrm{K}_{2} \mathrm{PtCl}_{4}$ with ammonia.

$$\mathrm{K}_{2} \mathrm{PtCl}_{4}(\mathrm{aq})+2 \mathrm{NH}_{3}(\mathrm{aq}) \longrightarrow \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(\mathrm{aq})+2 \mathrm{KCl}(\mathrm{aq})$$

If you begin with $16.0 \mathrm{g}$ of $\mathrm{K}_{2} \mathrm{PtCl}_{1},$ what mass of ammonia should be used to completely consume the $\mathrm{K}_{2} \mathrm{PtCl}_{4} ?$ What mass of cisplatin will be produced?

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okay, There's multiple parts to this question. The 1st 1 is simply determining mass percent based on the chemical Formula four CIS platinum with one mole platinum, two moles, nitrogen, six miles, hydrogen and two moles of chlorine. This, too, is distributed through the parentheses. So if we look up or calculate the molar mass assist, platinum is 301.1 grams per mole. So to determine the percent mass that is platinum, it'll be the Mat Moeller massive platinum multiplied by how many moles are present in one mole of CIS platinum, which is one divided by the molar mass of CIS. Platinum multiplied by 100 and we get 64.8% platinum. Then for nitrogen, it'll be the molar mass of nitrogen multiplied by how many moles of nitrogen are, insists platinum, which is too divided by the molar mass of CIS. Platinum multiplied by 100 and we get 9.30% nitrogen. For chlorine, it'll be the molar mass of chlorine multiplied by the moles of chlorine present, which is to insist platinum divided by the molar mass of CIS platinum multiplied by 100 and we get 25.4% chlorine. The rest if we sum up thes and subtract off. 100 will be the percent mass hydrogen to create CIS platinum if we take a two PT Seal four and reacted with two moles of ammonia, we do get CIS platinum according to the chemical reaction given in the question. So if we have 16 grams of this reactant, we can calculate the grant of ammonia required to react with it by simply dividing by the molar mass of Pete of K two, PTCL four. And then we can go from moles of this two moles of ammonia by multiplying by two and then moles of ammonia. Two grams of ammonia by multiplying by the molar mass of ammonia, which is 17.3 grams. And we get 1.31 grams of ammonia can be produced from 16 grams of K two PTCL four. Now we'll do a similar calculation to determine the massive CIS platinum in that can be created from the 16 grams of K two ptcl four and the 1.3 grams of ammonia that are required to react with it will convert the grams of this que to ptcl four toe moles by dividing by its Moeller Mass and will convert the moles of K two PTCL 42 moles of CIS platinum which is 1 to 1 based on the story Akiyama Tree of the Balanced Chemical Reaction and then go from moles of CIS platinum two grams of CIS platinum By multiplying by the molar mass of CIS platinum and we get 11.6 grams of CIS platinum can be produced.