Limiting Reactant Experiment PDF

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Limiting Reactant Experiment...

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Blank Blank March 1, 2017 Chem 131 Experiment #6: Limiting Reactant Purpose: Finding the limiting reactant in stoichiometric calculations is important because it lets the chemist/people working in the field of chemistry know what is happening in a chemical reaction (ex: how much of something is reacting). To find the limiting reactant of an unknown substance, filtration and digestion will be used. Filtration is the action or process of filtering something (Google dictionary) and digestion is the process of treating a substance by means of heat, enzymes, or a solvent to promote decomposition or extract essential components (Google dictionary). Adding separate samples of BaCl 2 and Na2PO4 to the two liquids is how a limited reactant was founded. Ions from the excess reactant must be carried to find the limiting reactant in the reaction. Since the mass of Ba3(PO4)2 is given, the mass of the limiting reactant could be found. Since the mass of the limiting reactant could be found, the percent by mass composition of the original mixture could also be found. Procedure: CHE 131 Experiment 6, General Chemistry I Lab, Winter Quarter 2016-2017. DePaul University. [Online] https://www.d2l.depaul.edu. (2/22/17). ● No changes were made to the previous procedures Data and Results Table 3: Suggested data table organization for the Sample Preparation The data in Table 3 is used to find the mass of the sample and percent composition determination. To find the mass of the initial sample and vial, and mass of the final sample and

vial is to weigh it on the scale. Make sure the scale is zeroed out before weighing anything to get the most accurate answer. To find the mass of sample: ● Mass of sample= mInitial- mFinal ● Mass of sample= 25.3757g- 23.6216g ● Mass of sample= 1.75 g These steps are repeated for Trial 2: ● Mass of sample: m=Initial-mFinal ● Mass of sample= 23.6216g-21.9852g ● Mass of sample= 1.64g Unknown#: 19 Trial 1

Trial 2

Limiting reactant (Ba2+ or PO4 3- )

BaCl2

BaCl2

Mass of initial sample and vial (g)

25.3757 g

23.6216

Mass of final sample and vial (g)

23.6216 g

21.9852 g

Mass of sample (g)

1.75g

1.64g

Table 4: Suggested Data Organization For Percent Composition Determination To find the mass of the filter and products, and mass of filter weigh the items on a scale. Make sure the scale is zeroed out before weighing the items to get the most accurate results. To find the mass of products: ● Mass of products= mFilter and product-mFilter

● Mass of product= 1.2384g-1.0234g ● Mass of product= 0.215 g Repeat the same steps for Trial 2: ● Mass of product= 1.3518g-1.0495g ● Mass of product= 0.302g Once the mass of the product is found, next is finding the moles of Ba3(PO4)2 product. This is done by dividing the mass of the product by the molar mass of Ba3(PO4)2. ● Moles of Ba3(PO4)2 = mProduct/mMolar mass ● Moles of Ba3(PO4)2 = 0.215g/601.924 g/mol ● Moles of Ba3(PO4)2 = 3.57*10^-4 Repeat the same steps for Trial 2: ● Moles of Ba3(PO4)2=0.302g/601.924 g/mol ● Moles of Ba3(PO4)2= 5.02*10^-4 Before doing any calculations to find the moles of limiting reactant, the dropper bottle test must be done to determine the limiting reactant in the unknown substance. The limiting reactant was Ba^2+. This was found by adding BaCl2 to the filtrate which formed a precipitate. ● Moles of limiting reactant= moles of Ba3(PO4)2 *(Ba^2+moles/ Ba3(PO4)2 moles ● Moles of limiting reactant= (3.57*10^-4 moles)*(4 mol of Ba^2+/1 mol of Ba3(PO4)2) ● Moles of limiting reactant= 1.42*10^-3 Repeat the same steps for Trial 2: ● Moles of limiting reactant= (5.02*10^-4 moles)* (4 mol of Ba^2+/ 1 mol of Ba3(PO4)2) ● Moles of limiting reactant= 2.01*10^-3

To find the mass of the limiting reactant, the moles of the limiting reactant should be multiplied by the molar mass of the limiting reactant ● Mass of limiting reactant=molLimiting reactant*molar massLimiting Reactant ● Mass of limiting reactant= 1.42*10^-3*137.327 ● Mass of limiting reactant= 0.195 Repeat steps for Trial 2 ● Mass of limiting reactant= molLimiting reactant*molar mass Limiting Reactant ● Mass of limiting reactant= 2.01*10^-3*137.327 ● Mass of limiting reactant= 0.276 To find the percent by mass of the limiting reactant is by dividing the limiting reactant mass by the mass of the original sample then multiplying by 100 ● % of BaCl2= (limiting reactant mass/original sample mass)*100 ● % of BaCl2= (0.195/0.215)*100 ● % of BaCl2=90.7% Repeat for Trial 2: ● % of BaCl2= (0.276/0.302)*100 ● % of BaCl2= 91.4% Trial 1

Trial 2

Mass of filter and product (g)

1.2384 g

1.3518g

Mass of filter (g)

1.0234 g

1.0495g

Mass of product (g)

0.215 g

0.302 g

Moles of Ba3(PO4)2 product (mol)

3.57*10^-4

5.02*10^-4

Moles of limiting reactant (mol)

1.42*10^-3

2.01*10^-3

Mass of limiting reactant (g)

0.195

0.276

% by mass BaCl2·2H2O

90.7%

91.4%

% by mass Na3PO4

9.3%

8.6%

Identification of Limiting Reaction and Percent Composition: The unknown number is #19. The limiting reactant is barium phosphate. The final average composition is in trial 1: 90.7%. The final average composition in trial 2: 91.4% Discussion The filtration process started with a solution in a beaker which needed to reach boiling temperature. After all the solute was poured into the filter, the precipitate must be washed with water while the remaining content was poured into the filter. The reason for this was to remove all of the leftover precipitate to the sides of the beaker to make sure the beaker was empty. Without doing this step the percent by mass of the limiting reagent would be low because of the residue left on the beaker, this is an important step in the procedure. To find the mass of the limiting reactant in trial 1, subtract the mass of the product from the mass of the limiting reactant. Excess reagent= mProduct- mLimiting reactant Excess reagent= 0.215 g- 0.195 g Excess reagent= 0.02 g...