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DedPeter [7]
3 years ago
13

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

carbon dioxide?
Chemistry
2 answers:
Mariulka [41]3 years ago
6 0
The answer on the quiz would be (d) 8.93 g 
julia-pushkina [17]3 years ago
3 0
Chemical equation of the reaction between citric acid and baking soda:

C_{6}  H_{8}  O_{7} +3NaHCO_{3} -\ \textgreater \  Na_{3} C_{6} H_{5} O_{7} +3 H_{2} O+3CO_{2}

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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Which of the following samples of baking soda wold react the fastest with
suter [353]
The answer is powder because if it was a small crystal it the molecules are tightly compact same with the small cube but there less compact, powder is loose and more spread out and easier to mix so it would react the fastest
3 0
3 years ago
If 8.500 g CH is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final
mojhsa [17]

The final temperature = 36 °C

<h3>Further explanation</h3>

The balanced combustion reaction for C₆H₆

2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l)  +6542 kJ

MW C₆H₆ : 78.11 g/mol

mol C₆H₆ :

\tt \dfrac{8.5}{78.11}=0.109

Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol

\tt \dfrac{0.109}{2}\times 6542=356.539~kJ/mol

Heat transferred to water :

Q=m.c.ΔT

\tt 356.539=5.691~kg\times 4.18~kj/kg^oC\times (t_2-21)\\\\t_2-21=15\rightarrow t_2=36^oC

3 0
3 years ago
CH4 + 202 → CO2 + 2H2O<br> How many grams of O2 needed to produce 36 grams of H2O?
Kipish [7]
<h3>Answer:</h3>

64 g O₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced]   CH₄ + 2O₂ → CO₂ + 2H₂O

[Given]   36 g H₂O

[Solve]   x g O₂

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol O₂ → 2 mol H₂O

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up conversion:                     \displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \  H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})
  2. Divide/Multiply [Cancel Units]:                                                                       \displaystyle 63.929 \ g \ O_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

63.929 g O₂ ≈ 64 g O₂

8 0
3 years ago
Read 2 more answers
Two closed vessels contain chlorine gas at the same conditions of temperature and
wariber [46]

Answer:12 mol

Explanation: both vessels are at the same temp and pressure (and the pressure is low and/or the temperature high).

6.7mol per 1.3L = 6.7/1.3 mol/L

so in 2.33L = 6.7*2.33/1.3 = 12 mol

8 0
3 years ago
True or Flase: dissolving salt in distilled water creates a homogeneous mixture.
chubhunter [2.5K]

Answer:

true

Explanation:

7 0
3 years ago
Read 2 more answers
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