I am struggling with the last step of this questions, balanced equation and % yi

Question

I am struggling with the last step of this questions, balanced equation and % yield.

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Name Date Preparation and Analysis of a Purple Nickel Compound Prelab 2.0184 g of a compound of the general formula Crx(NH3)yClz was prepared using 3.5246 CrCI3.6H20 and 10.00 mL of 15 MNH It was then analyzed as follows: Analysis of Cr 3 The concentration of cr was determined spectrophotometrically. The chromium ion is purple, with a maximum light absorption at 570 nm. A stock solution (Solution 1) of 0.100 M Cr" was used to make the following dilutions Solution 2: 10.00 mL stock solution diluted to a total of 100 mL with water c' Solution 3: 20.00 mL stock solution diluted to a total of 100 mL with water 1.2532 g of the unknown chromium compound was dissolved in enough water to make 100 mL of solution. Absorbance readings are given below: Absorbance Solution 1.850 00187 0.374 3 1.122 unknown compound calculate the concentration of Cr" in solutions 2 and 3. Then make a graph of absorbance vs. concentration using the three known (standard) solutions. From the graph, find the concentration of the chromium compound, calculate moles of Cr", grams of Cr 3, and Cr 3 in the unknown compound. Attach graph and show work below moles/L Cr" (from graph) Answers moles Cr g Cr 3 Cr

Explanation / Answer

Balance Equation is

CrCl3+3NH3 Cr (NH3)3 Cl3

calculate moles of CrCl3 & NH3

CrCl3 = 0.48 mol Cr + 1.44 mol of Cl =1.92 mol CrCl

NH3 =1.41 mol

from balance equation

1.92 mol of CrCl3 require 3x mol of NH3 i.e 1.92x 3 =5.76 mol NH3 but we have 1.41   mol NH3 hence it limiting reagent.

1.41mol NH3x(1mol Cr(NH3)3Cl3 / 3 mol NH3 ) = 0.47 mol of Cr(NH3)3Cl3

Convert into gm by multiplying molar mass of Cr(NH3)3Cl3

Cr(NH3)3Cl3 in gm =0.47 x 209.44 molar mass of Cr(NH3)3Cl3

Cr(NH3)3Cl3 in gm = 98.44 gm Theoretical yield

% yield = (2.0184 / 98.44 )x100 =2.05 %

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