Question

Last time I uploaded this someone said not enough info so I upload all the pages .can u please solve the questions with explanation so I can learn . Thank you

Chemistry 1050 Winter 2017-2018 EXPERIMENT 3 PREPARATION OF AN INORGANIC COMPOUND POTASSIUM TRISOXALATOFERRATEIII) TRIHYDRATE INTRODUCTION LEARNING OBJECTIVES To understand the tare method weighing using top-loading balance Preparation and isolation (using the method of crystallization) of an inorganic compound Caleulation of percent yield as an indication of reaction efficiency THEORY The reaction to be performed in the present experiment involves combining warm aqueous solutions of iron(IID) nitrate nonahydrate, potassium hydrogen oxalate, and potassium hydroxide to form the desired product, potassium trisoxalatoferrateam trihydrate (Ks(Fe(COhlaHro, hereafter referred to as "tris). (tris) Isolation of tris will be achieved by taking advantage of the differences in solubility of the produet in the reaction mixture at different temperatures and in different solvents. The solubility of substances is different different solvents or solvent mixtures. It is possible to begin the crystallization process of tris by adding s solvent mixture. Since tris has a lower solubility in ethanol than in water, it will begin ethanol to the aqueou crystallize from solution. Further crystallization can be achieved by taking advantage of the general observation that the solubility of most compounda decreases with a decrease in temperature. By placing to lask containing the partially crystallized product in an ice bath, thereby lowering its temperature, the solvent mixture is unable to solubilize as much tris and further crystallization of the product, from the cold mixture, results. TECHNIQUE VIDEOS TO WATCH Top-loading balance Weighing by difference Page 1 of 9

Answer 1

First question

The reaction given in the introduction is

(tris)

For the first reactant there are 4.2 grams (Molar mass = 404 g/gmol)

For the second reactant there are 3.8 grams (molar mass = 128.124 g/gmol)

The KOH is a 5ml solution 6 M

lets calculate the moles for each reactant

for Fe(NO3)*9H2O

n = mass / molar mass, where n is the number of moles

mass = moles * molar mass

molar mass = mass / moles

moles = 4.2 / 404 = 0.01039 moles

For the second reactant

moles = 3.8 / 128.124 = 0.0296 moles

moles of KOH

moles = Molarity * volume

moles = 6 * 0.005 = 0.03 moles

so in the reaction we need 1 mole of the first reactant

3 moles of the second reactant

3 moles of KOH

The analysis goes like this

1 mole of Fe(NO3)3*9H2O requires 3 moles of HKC2O4

we have 0.01039 of Fe(NO3)3*9H2O , this will require

3 * 0.01039 = 0.03117 moles of HKC2O4

and the same amount of KOH, 0.03117 moles

the amount of the second and the third reactant is less than the required so there is an excess of the first reactant Fe(NO3)3*9H2O  

With this analysis we can see that we have 0.0296 moles of HKC2O4 and we would need 0.03117, there are 0.03 moles of KOH so we can say that the HKC2O4 is the limiting reactant because the difference is bigger (existing vs required)

2. The theoretical yield calculation is made using the limiting reactant value, 3 moles of HKC2O4 produces 1 mole of tris

so the 0.0296 moles will produce

0.0296 / 3 = 0.00986667 moles of tris

the tris has a molar mass of 491.25 g/gmol

mass = moles * molar mass = 491.25 * 0.00986667 = 4.847 grams is the theoretical yield, this is the maximum amount that you could get

3.

I see you got 4.73 grams

The % yield is

4.73 / 4.847 * 100 = 97.58 %