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
Calculating the moles of water in 946.36 g of water=\frac{946.36 g}{236.59 g/mol}=4 mol=
236.59g/mol
946.36g
=4mol
Calculating the moles of sugar in 196.86 g of sugar=\frac{196.86 g}{225 g/mol}=0.8749 mol=
225g/mol
196.86g
=0.8749mol
Calculating the moles of lemon juice in 193.37 g of lemon juice=\frac{193.37 g}{257.83 g/mol}=0.7499 mol=
257.83g/mol
193.37g
=0.7499mol
Calculating the moles of lemonade in 2050.25 g of lemonade=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol=
719.42g/mol
2050.25g
=2.8498mol
It is apparent that the amount of lemon juice limits the reaction.
Consequently, we will base the reaction's completion on the quantity of lemon juice.
1 mole of lemon juice reacts with 2 moles of water, therefore, 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 utilized = 1.4998 mol × 236.59 g/mol=354.8376 g
Unused water = 946.36 g - 354.8376 g =591.5223 g
1 mole of lemon juice reacts with 1 mole of sugar, thus 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 sugar
Mass of sugar consumed = 0.7499 mol × 225 g/mol = 168.7275 g
Excess sugar = 196.86 g - 168.7275 g
1 mole of lemon juice results in 4 moles of lemonade.
Hence, 0.7499 mol of lemon juice will produce:
\frac{4}{1}\times 0.7499 mol=2.996 mol
1
4
×0.7499mol=2.996mol of lemonade
The mass of lemonade yielded = 2.996 mol × 719.42 g/mol = 2157.9722 g
Theoretical production of lemonade = 2157.9722 g
Actual yield of lemonade = 2050.25 g
Calculating the percentage yield of lemonade:
\frac{\text{Actual yield}}{\text{theoretical yield}}\times 100
theoretical yield
Actual yield
× 100
\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%
2157.9722g
2050.25g
× 100=95.00%
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The answer is a total of 74,844 calories.
Response:
of
can be found in 39.5 grams of
.
Clarification:
Atomic weights: P= 31, F= 19,
The molar mass equals 1 atomic weight of P + 5 atomic weights of
F= 31+5 × 19 = 31+95
=126 g/mole
The number of moles in 39.5 gm of
equals
=
=0.3134 moles
1 mole of any substance encompasses
0.3131 moles comprises 0.3134
Thus, of
can be found in 39.5 grams of
.