A 13.00 g sample of citric acid reacts with an excess of baking soda as shown in the equation. What is the theoretical yield of
2 answers:
Chemical equation of the reaction between citric acid and baking soda:
Ratios
Citric Acid Baking soda -> Sodium Citrate Water Carbon Dioxide
C6H8O7 CO2
1 mole 3moles
Molecular masses
Citric acid
6*12 + 8*1 + 7*16 = 192 g/mole
Carbon dioxide
12 + 2*16 = 44 g/mole
Proportion carbon dioxide / citric acid
3 * 44 g of carbon dioxide / 192 g of citric acid
132 g of carbon dioxide / 192 g of citric acid
13.00 g of citric acid * [132 g of carbon dioxide / 192 g of citric acid] = 8.94 g of carbon dioxide.
Answer: 8.94 grams.
Ratios
Citric Acid Baking soda -> Sodium Citrate Water Carbon Dioxide
C6H8O7 CO2
1 mole 3moles
Molecular masses
Citric acid
6*12 + 8*1 + 7*16 = 192 g/mole
Carbon dioxide
12 + 2*16 = 44 g/mole
Proportion carbon dioxide / citric acid
3 * 44 g of carbon dioxide / 192 g of citric acid
132 g of carbon dioxide / 192 g of citric acid
13.00 g of citric acid * [132 g of carbon dioxide / 192 g of citric acid] = 8.94 g of carbon dioxide.
Answer: 8.94 grams.
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Answer:
E°cell = - 1.3575 V
This reaction is spontaneous in the reverse direction
Explanation:
The given cell reaction:
Cr³⁺(1.32 × 10⁻³ M) + Fe²⁺(aq) → Cr²⁺(aq) + Fe³⁺(1.14 M)
The given Gibbs free energy: ΔG = 131 kJ = 131 × 10³ J (∵ 1 kJ = 10³ J)
As we know,
ΔG = - n F E°cell
Here, n - the number of moles of electrons transferred = 1
F - Faraday constant = 96500
E°cell - cell potential = ?
<u>For a given chemical reaction if-</u>
1. ΔG = negative and E°cell = positive
⇒ <em>The reaction is spontaneous and proceeds spontaneously in the forward direction.</em>
2. ΔG = positive and E°cell = negative
⇒ <em>The reaction is non-spontaneous and proceeds spontaneously in the reverse direction.</em>
<u>Since, for this chemical reaction: </u>
Cr³⁺(1.32 × 10⁻³ M) + Fe²⁺(aq) → Cr²⁺(aq) + Fe³⁺(1.14 M)
ΔG = + 131 × 10³ J ⇒ positive
and, E°cell = - 1.3575 V ⇒ negative
<u>Therefore, this reaction is spontaneous in the reverse direction.</u>
<u>Answer:</u> The molar mass of unknown triprotic acid is 97.66 g/mol
<u>Explanation:</u>
To calculate the molarity of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of triprotic acid
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:
We are given:
Molarity of solution = 0.0077 M
Given mass of triprotic acid = 0.188 g
Volume of solution = 250 mL
Putting values in above equation, we get:
Hence, the molar mass of unknown triprotic acid is 97.66 g/mol