All categories

2 years ago

Just Lemons Lemonade Recipe Equation:

Chemistry

1 answer:

zalisa [80]2 years ago

7 0

Answer:

Explanation:

Hello!

Complete text:

Honors Stoichiometry Activity WorksheetInstructions:

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 946.36 g

Sugar 196.86 g

Lemon Juice 193.37 g

Lemonade 2050.25g

Leftover Ingredients?

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 = 225 g

1 mole of lemon juice = 1 cup = 257.83 g

1 mole of lemonade = 1 cup = 719.42 g

You have the information on the ingredients used to produce one batch of lemonade and the amount of lemonade produced. To determine which ingredients be leftovers, you have to determine first, which one is the limiting reactant, i.e. the ingredient that will be used up first.

According to the recipe, to make 4 moles of lemonade, you use 2 moles of water, one mole of sugar and one mole of lemon juice, expressed in grams:

2 water + sugar + lemon juice = 4 lemonade

2*(236.59) + 225g + 257.83g = 4*(719.42)g

473.18g + 225g + 257.83g = 2877.68g

So for every 2877.68g of lemonade made, they use 473.18g of water, 225g of sugar, and 257.83g of lemon juice.

You know that they made a batch of 2050.25g, so to detect the limiting reactant, first, you have to calculate, in theory, how much of each ingredient you need to make the given amount of lemonade:

Use cross multiplication

Water:

2877.68g lemonade → 473.18g water

2050.25g lemonade → X= (2050.25*473.18)/2877.68= 337.12g water

Following the recipe, to elaborate 2050.25g of lemonade, you need to use 337.12g of water.

Sugar:

2877.68g lemonade → 225g sugar

2050.25g lemonade → X= (2050.25*225)/2877.68= 160.30g sugar.

To elaborate 2050.25f of lemonade you need to use 160.30g of sugar.

Lemon juice:

2877.68g lemonade → 257.83g lemon juice

2050.25g lemonade → X= (2050.25*257.83)/2877.68= 183.69g lemon juice.

To elaborate 2050.25f of lemonade you need to use 183.69g lemon juice.

Available ingredients vs. theoretical yields for 2050.25g of lemonade:

Water 946.36 g → 337.12g

Sugar 196.86 g → 160.30g

Lemon Juice 193.37 g → 183.69g

The lemon juice will be the first ingredient to be used up, there will be a surplus of water and sugar.

I hope this helps!

You might be interested in

A steam turbine is a device in which high-pressure steam enters the turbine and causes blades to spin. The spinning blades have

pav-90 [236]

The answer is Some of the heat used to make the steam is transformed into work as the steam pushes against the blades to make them turn. This also observes the law of conservation in thermodynamics. The kinetic energy in the steam helps push the blades and is converted to mechanical energy. Some of the energy is lost since it is never 100% efficient.

6 0

2 years ago

Read 2 more answers

The name "penicillin" is used for several closely-related antibiotics. A 1.2177 g sample of one of these compounds is burned, pr

masya89 [10]

Answer: The empirical formula for the given organic compound is $C_7H_{11}O_2SN$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$S_vN_wC_xH_yO_z+O_2\rightarrow CO_2+H_2O+SO_2+NO$

where, 'v', 'w' 'x', 'y' and 'z' are the subscripts of sulfur, nitrogen, carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=0.2829g$

Mass of $H_2O=0.1159g$

Mass of $SO_2=0.4503g$

Mass of NO = 0.2109 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

Molar mass of sulfur dioxide = 64 g/mol

Molar mass of nitrogen monoxide = 30 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 2.1654 g of carbon dioxide, $\frac{12}{44}\times 2.1654=0.590g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.6965 g of water, $\frac{2}{18}\times 0.6965=0.077g$ of hydrogen will be contained.

For calculating the mass of sulfur:

In 64 g of sulfur dioxide, 32 g of sulfur is contained.

So, in 0.4503 g of sulfur dioxide, $\frac{32}{64}\times 0.4503=0.225g$ of sulfur will be contained.

For calculating the mass of nitrogen:

In 30 g of nitrogen monoxide, 14 g of nitrogen is contained.

So, in 0.2109 g of nitrogen monoxide, $\frac{14}{30}\times 0.2109=0.098g$ of nitrogen will be contained.

Mass of oxygen in the compound = (1.2177) - (0.590 + 0.077 + 0.225 + 0.098) = 0.2277 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.590g}{12g/mole}=0.049moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.077g}{1g/mole}=0.077moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.2277g}{16g/mole}=0.0142moles$

Moles of Sulfur = $\frac{\text{Given mass of Sulfur}}{\text{Molar mass of sulfur}}=\frac{0.225g}{32g/mole}=0.007moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{0.098g}{14g/mole}=0.007moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.007 moles.

For Carbon = $\frac{0.049}{0.007}=7$

For Hydrogen = $\frac{0.077}{0.007}=11$

For Oxygen = $\frac{0.0142}{0.007}=2.03\approx 2$

For Sulfur = $\frac{0.007}{0.007}=1$

For Nitrogen = $\frac{0.007}{0.007}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O : S : N = 7 : 11 : 2 : 1 : 1

Hence, the empirical formula for the given compound is $C_7H_{11}O_2S_1N_1=C_7H_{11}O_2SN$

4 0

2 years ago

How long will it take 10.0 mL of Ne gas to effuse through a porous barrier if it has been observed that 125 minutes are required

cupoosta [38]

Answer:

88.8 minutes

Explanation:

Graham's law of diffusion relates rate of difusion by the following formula

Rate1 / rate 2 = √( Mass of argon / Mass of Neon)

Where rate = volume divided by time

Rate 1 = 10 ml / t1

Rate 2 = 10 ml / t2

Rate 1/ rate 2 = 10 ml / t1 ÷ 10 ml/ t2 = t2/ t1

t2/t1 = √(Mass of argon / mass of Neon) = √( 39.984/20.179)

125 / t1 = 1.4026

t1 = 125 / 1.4026 = 88.8 minutes

7 0

2 years ago

A 7.591-9 gaseous mixture contains methane (CH4) and butane

mestny [16]

Answer:

65.71%

Explanation:

First, we can write the mass of the mixture, thus:

7.519g = X + Y (1)

Where X is the mass of methane and Y the mass of butane

Also, the reactions of combustion are:

CH₄ + 2O₂ → CO₂ + 2H₂O

2 moles of oxygen react per mole of methane

C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O

13/2 moles of oxygen react per mole of methane

That means, in therms of moles of oxygen we can write:

0.9050 moles = 2X/16.04 + 13/2Y/ 58.12

0.9050 = 0.12469X + 0.11184Y (2)

Where 16.04 and 58.12 are molar masses of methane and butane

That is because if X is the mass of methane:

X g Methane * (1mol / 16.04g) = Moles methane

Moles methane * (2 moles Oxygen / mole methane) = Moles oxygen

Replacing (1) in (2):

0.9050 = 0.12469X + 0.11184 (7.519 - X)

0.9050 = 0.12469X + 0.841 - 0.11184X

0.0641 = 0.01285X

X = 4.988g = Mass of methane.

And mass percent of methane is:

4.988g / 7.591g * 100

7 0

2 years ago

The decantate from Question 1 that possibly contains K, Zn, or Cr, is neutralized with acetic acid and divided into the appropri

stepladder [879]

Answer:

2 that is Calcium (Ca) and Manganese (Mn)

Explanation:

Add dithizone in chloroform and look for a red complex formation.

4 0

2 years ago