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3 years ago

Given the following equation: CH4 + 202 CO2 + 2H2O

Chemistry

2 answers:

trapecia [35]3 years ago

8 0

Answer:

Mass = 0.8 g

Explanation:

Given data:

Mass of oxygen = 3.2 g

Mass of CH₄ = ?

Solution:

Chemical equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

Number of moles of O₂:

Number of moles = mass/molar mass

Number of moles = 3.2 g/ 32 g/mol

Number of moles = 0.1 mol

Now we will compare the moles of CH₄ and O₂.

O₂ : CH₄

2 : 1

0.1 : 1/2 = 0.1 = 0.05

Mass of CH₄:

Mass = number of moles × molar mass

Mass = 0.05 mol × 16 g/mol

Mass = 0.8 g

lawyer [7]3 years ago

4 0

Answer:

0.8

Explanation:

Chunga

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A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa

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Answer: The identity of the unknown acid is butanoic acid or ascorbic acid.

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To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$

Molarity of NaOH solution = 0.570 M

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Putting values in above equation, we get:

$0.570M=\frac{\text{Moles of NaOH}\times 1000}{39.55}\\\\\text{Moles of NaOH}=\frac{0.570\times 39.55}{1000}=0.0225mol$

The chemical equation for the reaction of NaOH and monoprotic acid follows:

$NaOH+HX\rightarrow NaX+H_2O$

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HX

So, moles of monoprotic acid = 0.0225 moles

The chemical equation for the reaction of NaOH and diprotic acid follows:

$2NaOH+H_2X\rightarrow 2NaX+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

So, moles of diprotic acid = $\frac{0.0225}{2}=0.01125moles$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For butanoic acid:

Mass of butanoic acid = 2.002 g

Molar mass of butanoic acid = 88 g/mol

Putting values in above equation, we get:

$\text{Moles of butanoic acid}=\frac{2.002g}{88g/mol}=0.02275mol$

For L-tartaric acid:

Mass of L-tartaric acid = 2.002 g

Molar mass of L-tartaric acid = 150 g/mol

Putting values in above equation, we get:

$\text{Moles of L-tartaric acid}=\frac{2.002g}{150g/mol}=0.0133mol$

For ascorbic acid:

Mass of ascorbic acid = 2.002 g

Molar mass of ascorbic acid = 176 g/mol

Putting values in above equation, we get:

$\text{Moles of ascorbic acid}=\frac{2.002g}{176g/mol}=0.01137mol$

As, the number of moles of butanoic acid and ascorbic acid is equal to the number of moles of acid getting neutralized.

Hence, the identity of the unknown acid is butanoic acid or ascorbic acid.

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