100 POINTS PLEASE HELP!! Honors Stoichiometry Activity Worksheet Instructions: In this laboratory activity, you will taste test
two samples of Just Lemons lemonade for taste quality. Then you will analyze lemonade production data for percent yield and excess ingredients. Complete each section of this worksheet, and submit it to your instructor for grading. Activity One: Tasting Excess and Limiting Ingredients Make lemonade samples using the sample 1 and sample 2 recipes. These represent two different batches of Just Lemons lemonade. Record your taste observations for each sample in the data chart. Sample 1: 1 cup water 2/3 cup sugar 1/2 cup lemon juice Sample 2: 1 cup water 1/2 cup sugar 1/4 cup lemon juice Taste Observations Sample 1: Sample 2: Activity Two: Just Lemons, Inc. Production Here's a one-batch sample of Just Lemons lemonade production. Determine the percent yield and amount of leftover ingredients for lemonade production and place your answers in the data chart. Hint: Complete stoichiometry calculations for each ingredient to determine the theoretical yield. Complete a limiting reactant-to-excess reactant calculation for both excess ingredients. Water Sugar Lemon Juice Lemonade Percent Yield Leftover Ingredients 946.36 g 196.86 g 193.37 g 719.84 g Just Lemons Lemonade Recipe Equation: 2 water + sugar + lemon juice = 4 lemonade Mole conversion factors: 1 mole of water = 1 cup = 236.59 g 1 mole of sugar = 1 cup = 198 g 1 mole of lemon juice = 1 cup = 229.96 g 1 mole of lemonade = 1 cup = 225.285 g Show your calculations below. Analysis Questions 1. Based on taste observations only, which ingredients were in excess in the lemonade samples in Activity One? 2. Based on the data in Activity Two, which excess ingredients are affecting the taste of the lemonade in the sample batch? 3. What can Just Lemons, Inc. do during production to reduce the amount of excess ingredients and improve the taste of their lemonade? 4. Try to reduce the amount of leftover ingredients by changing the amount of one, two, or all three starting ingredients. Show your stoichiometric calculations below. 5. During factory inspection, Just Lemons, Inc. discovered that a water valve to the lemonade mixing station was not functioning. Once they repair it, more water will enter the mixing station. From what you know about the limiting and excess ingredients for current lemonade production, what advice would you give engineers about the upcoming increase in water?
2 answers:
Answer:
2 water + sugar + lemon juice → 4 lemonade
Moles of water present in 946.36 g of water=\frac{946.36 g}{236.59 g/mol}=4 mol=
236.59g/mol
946.36g
=4mol
Moles of sugar present in 196.86 g of water=\frac{196.86 g}{225 g/mol}=0.8749 mol=
225g/mol
196.86g
=0.8749mol
Moles of lemon juice present in 193.37 g of water=\frac{193.37 g}{257.83 g/mol}=0.7499 mol=
257.83g/mol
193.37g
=0.7499mol
Moles of lemonade in 2050.25 g of water=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol=
719.42g/mol
2050.25g
=2.8498mol
As we can see that number of moles of lemon juice are limited.
So, we will consider the reaction will complete in accordance with moles of lemon juice.
1 mole lemon juice reacts with 2 mol of water,then 0.7499 mol of lemon juice will react with:
\frac{2}{1}\times 0.7499 mol = 1.4998 mol
1
2
×0.7499mol=1.4998mol of water
Mass of water used = 1.4998 mol × 236.59 g/mol=354.8376 g
Water remained unused = 946.36 g - 354.8376 g =591.5223 g
1 mole lemon juice reacts with mol of sugar,then 0.7499 mol of lemon juice will react with:
\frac{1}{1}\times 0.7499 mol = 0.7499 mol
1
1
×0.7499mol=0.7499mol of water
Mass of sugar used = 0.7499 mol × 225 g/mol = 168.7275 g
Sugar remained unused = 196.86 g - 28.1325 g
1 mole of lemon juice gives 4 moles of lemonade.
Then 0.7499 mol of lemon juice will give:
\frac{4}{1}\times 0.7499 mol=2.996 mol
1
4
×0.7499mol=2.996mol of lemonade
Mass of lemonade obtained = 2.996 mol × 719.42 g/mol = 2157.9722 g
Theoretical yield of lemonade = 2157.9722 g
Experimental yield of lemonade = 2050.25 g
Percentage yield of lemonade:
\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100
theoretical yield
Experimental yield
×100
\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%
2157.9722g
2050.25g
×100=95.00%
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Answer:
Aluminum metal
Explanation:
In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.
First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:
Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.
Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.
Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive):
Notice that the overall cell potential upon summing is:
Meaning we obey the law of galvanic cells.
Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.