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user100 [1]
3 years ago
7

CH4 + 202 → CO2 + 2H2O How many grams of O2 produced from 2 moles of H20?

Chemistry
2 answers:
Snezhnost [94]3 years ago
8 0
<h3>Answer:</h3>

60 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]   2 mol H₂O

[Solve]   x g O₂

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

[RxN] 2 mol H₂O → 2 mol O₂

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

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

<u>Step 3: Stoichiometry</u>

  1. Set up conversion:                    \displaystyle 2 \ mol \ H_2O(\frac{2 \ mol \ O_2}{2 \ mol \ H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})
  2. Divide/Multiply:                          \displaystyle 64.00 \ g \ O_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

64.00 g O₂ ≈ 60 g O₂

Finger [1]3 years ago
4 0

Mass of O₂= 64 g

<h3> Further explanation </h3>

Given

Reaction

CH₄ + 20₂ → CO₂ + 2H₂O

2 moles of H₂O

Required

Mass of O₂

Solution

In a chemical equation, the reaction coefficient shows the mole ratio of the compounds involved in the reaction, reactants or products

From the equation, mol ratio of H₂O  : O₂ = 1 : 1, so moles O₂ :

= 1/1 x moles H₂O

= 1/1 x 2

= 2 moles

Mass of O₂ :

= mol x MW

= 2 x 32 g/mol

= 64 g

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Answer:

Both plants and animals release energy from glucose in cellular respiration.

4 0
3 years ago
217 mL of neon gas at 551 celsius is brought to standard temperature (0 celsius) while the pressure is held constant. What is th
pantera1 [17]

Answer: The new volume is 72 ml

Explanation:

To calculate the final volume of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=217ml\\T_1=551^oC=(551+273)K=824K\\V_2=?\\T_2=0^0C=(0+273)K=273K

Putting values in above equation, we get:

\frac{217ml}{824K}=\frac{V_2}{273K}\\\\V_2=72ml

Thus the new volume is 72 ml

7 0
2 years ago
I have a chemistry question that I am needing help on please?​
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What country and grade are you?
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8 0
3 years ago
A reaction generates hydrogen gas (H) as a product. The re-
Amiraneli [1.4K]

<u>Answer:</u> The average rate of the reaction is 7.82\times 10^{-3}M/min

<u>Explanation:</u>

To calculate the molarity of hydrogen gas generated, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of hydrogen gas = 3.91\times 10^{-2}mol

Volume of solution = 250 mL = 0.250 L    (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

\text{Molarity of }H_2=\frac{3.91\times 10^{-2}mol}{0.250L}=0.1564M

Average rate of the reaction is defined as the ratio of concentration of hydrogen generated to the time taken.

To calculate the average rate of the reaction, we use the equation:

\text{Average rate of the reaction}=\frac{\text{Concentration of hydrogen generated}}{\text{Time taken}}

We are given:

Concentration of hydrogen generated = 0.1564 M

Time taken = 20.0 minutes

Putting values in above equation, we get:

\text{Average rate of the reaction}=\frac{0.1564M}{20.0min}\\\\\text{Average rate of the reaction}=7.82\times 10^{-3}M/min

Hence, the average rate of the reaction is 7.82\times 10^{-3}M/min

6 0
2 years ago
"An aqueous CaCl2 solution has a vapor pressure of 83.1mmHg at 50 ∘C. The vapor pressure of pure water at this temperature is 92
Lynna [10]

Answer : The the concentration of CaCl_2 in mass percent is, 41.18 %

Solution : Given,

Molar mass of water = 18 g/mole

Molar mass of CaCl_2 = 110.98 g/mole

First we have to calculate the mole fraction of solute.

According to the relative lowering of vapor pressure, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component of the solution multiplied by the vapor pressure of that component in the pure state.

\fac{p^o-p_s}{p^o}=X_B

where,

p^o = vapor pressure of the pure component (water) = 92.6 mmHg

p_s = vapor pressure of the solution = 83.1 mmHg

X_B = mole fraction of solute, (CaCl_2)

Now put all the given values in this formula, we get the mole fraction of solute.

\fac{92.6-83.1}{92.6}=X_B

X_B=0.102

Now we have to calculate the mole fraction of solvent (water).

As we know that,

X_A+X_B=1\\\\X_A=1-X_B\\\\X_A=1-0.102\\\\X_A=0.898

The number of moles of solute and solvent will be, 0.102 and 0.898 moles respectively.

Now we have to calculate the mass of solute, (CaCl_2) and solvent, (H_2O).

\text{Mass of }CaCl_2=\text{Moles of }CaCl_2\times \text{Molar mass of }CaCl_2

\text{Mass of }CaCl_2=(0.102mole)\times (110.98g/mole)=11.32g

\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O

\text{Mass of }H_2O=(0.898mole)\times (18g/mole)=16.164g

Mass of solution = Mass of solute + Mass of solvent

Mass of solution = 11.32 + 16.164 = 27.484 g

Now we have to calculate the mass percent of CaCl_2

Mass\%=\frac{\text{Mass of}CaCl_2}{\text{Mass of solution}}\times 100=\frac{11.32g}{27.484g}\times 100=41.18\%

Therefore, the the concentration of CaCl_2 in mass percent is, 41.18 %

4 0
2 years ago
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