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Brums [2.3K]
3 years ago
5

2. Suppose that 21.37 mL of NaOH is needed to titrate 10.00 mL of 0.1450 M H2SO4 solution.

Chemistry
1 answer:
AysviL [449]3 years ago
7 0

Answer:

0.1357 M

Explanation:

(a) The balanced reaction is shown below as:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

(b) Moles of H_2SO_4 can be calculated as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For H_2SO_4 :

Molarity = 0.1450 M

Volume = 10.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 10×10⁻³ L

Thus, moles of H_2SO_4 :

Moles=0.1450 \times {10\times 10^{-3}}\ moles

Moles of H_2SO_4  = 0.00145 moles

From the reaction,

1 mole of H_2SO_4 react with 2 moles of NaOH

0.00145 mole of H_2SO_4 react with 2*0.00145 mole of NaOH

Moles of NaOH = 0.0029 moles

Volume = 21.37 mL = 21.37×10⁻³ L

Molarity = Moles / Volume = 0.0029 /  21.37×10⁻³  M = 0.1357 M

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Answer:

Theoretical yield = 3.51 g

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

For CH_4  :-

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Molar mass of CH_4  = 16.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

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For O_2  :-

Mass of O_2  = 10.1 g

Molar mass of O_2  = 31.998 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

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Moles= \frac{10.1\ g}{31.998\ g/mol}

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According to the given reaction:

CH_4+2O_2\rightarrow CO_2+2H_2O

1 mole of methane gas reacts with 2 moles of oxygen gas

0.0798 mole of methane gas reacts with 2*0.0798 moles of oxygen gas

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<u>Limiting reagent is the one which is present in small amount. Thus, CH_4 is limiting reagent. </u>

The formation of the product is governed by the limiting reagent. So,

1 mole of methane gas on reaction produces 1 mole of carbon dioxide.

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moles = \frac{Mass\ taken}{Molar\ mass}

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3 years ago
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