3. Given the following equation: underline ( )K_(2)PtCl_(4)+underline ( )NH_(3) arrow underline (2)Pt(NH_(3))_(2)Cl_(2)+underline ( )KCl a) Determine the theoretical yield of KCU you start with 34.5 grams of NH3 Answer: __ b) Starting with 64.5g of NH3 and you actually obtain 86.4gof Pt(NH_(3))_(2)Cl_(2) what is the percent yield __

a) To determine the theoretical yield of KCl, we first need to balance the chemical equation. The balanced equation is:<br /><br />\[ K_2PtCl_4 + 2NH_3 \rightarrow 2Pt(NH_3)_2Cl_2 + 2KCl \]<br /><br />From the balanced equation, we can see that 2 moles of NH3 produce 2 moles of KCl. The molar mass of NH3 is 17.03 g/mol, and the molar mass of KCl is 74.55 g/mol.<br /><br />First, we calculate the moles of NH3:<br /><br />\[ \text{Moles of NH3} = \frac{\text{Mass of NH3}}{\text{Molar mass of NH3}} = \frac{34.5 \text{ g}}{17.03 \text{ g/mol}} \approx 2.03 \text{ moles} \]<br /><br />Since the reaction is 1:1, the moles of KCl produced will be the same as the moles of NH3 used. Now we calculate the mass of KCl produced:<br /><br />\[ \text{Mass of KCl} = \text{Moles of KCl} \times \text{Molar mass of KCl} = 2.03 \text{ moles} \times 74.55 \text{ g/mol} \approx 151.5 \text{ g} \]<br /><br />So, the theoretical yield of KCl is 151.5 grams.<br /><br />b) To calculate the percent yield, we use the formula:<br /><br />\[ \text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100\% \]<br /><br />The theoretical yield of \( Pt(NH_3)_2Cl_2 \) can be calculated using the stoichiometry of the reaction. From the balanced equation, 2 moles of NH3 produce 2 moles of \( Pt(NH_3)_2Cl_2 \). The molar mass of \( Pt(NH_3)_2Cl_2 \) is 237.73 g/mol.<br /><br />First, we calculate the moles of \( Pt(NH_3)_2Cl_2 \) produced:<br /><br />\[ \text{Moles of } Pt(NH_3)_2Cl_2 = \frac{\text{Mass of } Pt(NH_3)_2Cl_2}{\text{Molar mass of } Pt(NH_3)_2Cl_2} = \frac{86.4 \text{ g}}{237.73 \text{ g/mol}} \approx 0.363 \text{ moles} \]<br /><br />Now we calculate the theoretical yield of \( Pt(NH_3)_2Cl_2 \):<br /><br />\[ \text{Theoretical yield of } Pt(NH_3)_2Cl_2 = \text{Moles of } Pt(NH_3)_2Cl_2 \times \text{Molar mass of } Pt(NH_3)_2Cl_2 = 0.363 \text{ moles} \times 237.73 \text{ g/mol} \approx 86.4 \text{ g} \]<br /><br />Now we can calculate the percent yield:<br /><br />\[ \text{Percent yield} = \left( \frac{86.4 \text{ g}}{86.4 \text{ g}} \right) \times 100\% = 100\% \]