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3 years ago

13

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 2050.25 g

Could you please figure out the percent yield and the
leftover ingredients

1 answer:

Bumek [7]3 years ago

4 0

the percent yield is 95% and the excess ingredients are sugar and water but i don't know the amounts. pls let me know if you do :(

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Which processes occur when energy is removed from a substance

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Answer:

Note that melting and vaporization are endothermic processes in that they absorb or require energy, while freezing and condensation are exothermic process as they release energy.

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2 years ago

A laboratory analysis of a sample finds it is composed of 38.8% carbon, 16.2% hydrogen, and 45.1% nitrogen. What is its empirica

Sladkaya [172]

Answer: The empirical formula for the given compound is $CH_5N$

Explanation : Given,

Percentage of C = 38.8 %

Percentage of H = 16.2 %

Percentage of N = 45.1 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 38.8 g

Mass of H = 16.2 g

Mass of N = 45.4 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{38.8g}{12g/mole}=3.23moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{16.2g}{1g/mole}=16.2moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{45.4g}{14g/mole}=3.24moles$

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 3.23 moles.

For Carbon = $\frac{3.23}{3.23}=1$

For Hydrogen = $\frac{16.2}{3.23}=5.01\approx 5$

For Oxygen = $\frac{3.24}{3.23}=1.00\approx 1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 1 : 5 : 1

Hence, the empirical formula for the given compound is $C_1H_5N_1=CH_5N$

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3 years ago

Hydrocyanic acid, HCN, is a weak acid. (a) Write the chemical equation for the dissociation of HCN in water. (b) Identify the Br

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Answer: a) $HCN(aq.)+H_2O\rightleftharoons H_3O^+(aq.)+CN^-(aq.)$

b) $HCN$ : acid $CN^-$ :conjugate base.

And, $H_2O$ : base $H_3O^+$ : conjugate acid.

c) $HCN(aq.)+NaOH(aq)\rightleftharoons NaCN(aq.)+H_2O(l)$

d) $NaCN(aq)\rightarrow Na^+(aq.)+CN^-(aq)$

e) $NaCN(aq)+HCl(aq)\rightarrow NaCl(aq.)+HCN(aq)$

Explanation:

a) Weak acid is defined as the acid which does not completely dissociates when dissolved in water. They have high pH. These releases $H^+$ ions in their aqueous states.

The equation for the dissociation of $HCN$ acid is given by:

$HCN(aq.)+H_2O\rightleftharoons H_3O^+(aq.)+CN^-(aq.)$

b) According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

For the given chemical equation:

$HCN$ is loosing a proton, thus it is considered as an acid and after losing a proton, it forms $CN^-$ which is a conjugate base.

And, $H_2O$ is gaining a proton, thus it is considered as a base and after gaining a proton, it forms $H_3O^+$ which is a conjugate acid.

c) Neutralization reaction is a reaction in which an acid reacts with base to produce salt and water.

$HCN(aq.)+NaOH(aq)\rightarrow NaCN(aq.)+H_2O(l)$

d) The chemical equation for dissociation of $NaCN$ in water.

$NaCN(aq)\rightarrow Na^+(aq.)+CN^-(aq)$

e) The chemical equation for the reaction of $NaCN$ and $HCI$

$NaCN(aq)+HCl(aq)\rightarrow NaCl(aq.)+HCN(aq)$

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3 years ago

Determine the concentration of H+ in each solution at 25∘C

puteri [66]

Answer:

1: [H+] = 0.01 M

2: [H+] = 0.0001 M

3: [H+] = 0.0001 M

Explanation:

Step 1: data given

pH = -log[H+]

pH = pOH = 14

Step 2:

1. A solution with pH = 2.0

pH = 2

-log[H+] = 2.0

[H+] = 10^-2

[H+] = 0.01 M

2. A solution with pH = 4.0

pH = 4

-log[H+] = 4.0

[H+] = 10^-4

[H+] = 0.0001 M

3. A solution with pOH = 10.0

pH = = 14 - 10 = 4

pH = 4

-log[H+] = 4.0

[H+] = 10^-4

[H+] = 0.0001 M

4 0

3 years ago

Calculate the theoretical yield of alum expected from 0.9875 g of aluminum foil. assume the aluminum is the limiting reactant.

skelet666 [1.2K]

Answer: 17.34 grams of alum will be produced if 0.9875 g of Aluminium foil was used.

Explanation: Reaction to form alum from Aluminium is given as:

$2Al(s)+2KOH(aq.)+2H_2O(l)+4H_2SO_4(aq.)\rightarrow 2KAl(SO_4)_2(s).12H_2O(l)+3H_2(g)$

We are given Aluminium to be the limiting reactant, so the formation of alum will be dependent on Aluminium because it limits the formation of product.

By stoichiometry,

2 moles of Al is producing 2 moles of Alum

Mass of 2 moles of Aluminium = (2 × 27)g/mol = 54 g/mol

Mass of 2 moles of alum = (2 × 474)g/mol = 948 g/mol

54 g/mol of aluminium will produce 948 g/mol of alum, so

$\text{0.9875 grams of aluminium will produce}=\frac{948g/mol}{54g/mol}\times 0.9875g$

Amount of Alum produced = 17.34 grams

Theoretical yield of alum = 17.34 grams.

5 0

3 years ago