1K2PtCl4+2(NH3)->1Pt(NH3)2Cl2+2KCl

what is the percent yield, when 34.5 grams of NH3 react with excess K2PtCl4 if 76.4 g of Pt(NH3)2Cl2 are actually produced?

See your Al2(SO4)3 problem.

To calculate the percent yield, use the following formula:

Percent Yield = (Actual Yield/Theoretical Yield) x 100

In this chemical reaction, the reactant NH3 is limiting, which means it determines the maximum amount of product that can be formed. Therefore, we need to calculate the theoretical yield of Pt(NH3)2Cl2 using the stoichiometry of the reaction.

First, let's calculate the moles of NH3 using its molar mass (17.03 g/mol):

Moles of NH3 = Mass of NH3 / Molar mass of NH3
Moles of NH3 = 34.5 g / 17.03 g/mol
Moles of NH3 = 2.027 mol NH3

According to the balanced chemical equation, the stoichiometry ratio between NH3 and Pt(NH3)2Cl2 is 2:1. Therefore, the moles of Pt(NH3)2Cl2 produced will be half of the moles of NH3:

Moles of Pt(NH3)2Cl2 = Moles of NH3 / 2
Moles of Pt(NH3)2Cl2 = 2.027 mol NH3 / 2
Moles of Pt(NH3)2Cl2 = 1.014 mol Pt(NH3)2Cl2

Next, let's calculate the theoretical yield of Pt(NH3)2Cl2 using its molar mass (413.9 g/mol):

Theoretical Yield = Moles of Pt(NH3)2Cl2 * Molar mass of Pt(NH3)2Cl2
Theoretical Yield = 1.014 mol * 413.9 g/mol
Theoretical Yield = 419.47 g Pt(NH3)2Cl2

Now, we can calculate the percent yield:

Percent Yield = (Actual Yield / Theoretical Yield) x 100
Percent Yield = (76.4 g / 419.47 g) x 100
Percent Yield = 18.22%

So, the percent yield of Pt(NH3)2Cl2 in this reaction is approximately 18.22%.