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2 years ago

3. At STP, how many liters of oxygen gas are required to react completely with 3.2 x 10^22

Chemistry

1 answer:

Mrrafil [7]2 years ago

7 0

The volume of oxygen required to react with 3.2 * 10²² molecules of hydrogen is 0.56 L.

<h3>What is stoichiometric law?</h3>

The stoichiometric law has been given as the representation of the moles of product and reactant in a chemical reaction are represented by the stoichiometric coefficient.

It has been known that 1 mole of a compound has 6.023 * 10²³ molecules. Thus, the moles of hydrogen gas equivalent to 3.2 * 10²² molecules has been:

6.023 * 10²³ molecules = 1 mole

3.2 * 10²² molecules = 1/6.023 * 10²³ * 3.2 * 10²² moles

3.2 * 10²² molecules = 0.05 moles

Thus, the moles of hydrogen gas available is 0.05 moles.

From the stoichiometric law, according to the balanced chemical equation,

2 moles of hydrogen requires = 1 moles of oxygen

0.05 moles of hydrogen requires = 1/2 * 0.05 moles oxygen

0.05 moles of hydrogen requires = 0.025 moles oxygen

Thus, the moles of oxygen required is 0.025 moles. At STP, a mole of gas been equivalent to 22.4 L . Thus, the volume of 0.025 moles oxygen has been:

1 mole = 22.4 L

0.025 moles = 22.4 * 0.025 L

0.025 mole = 0.56 L

Thus, the volume of oxygen required to react with 3.2 * 10²² molecules of hydrogen is 0.56 L.

Learn more about stoichiometric law, here:

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Answer: The empirical and molecular formula for the given organic compound are $CH$ and $C_2H_2$ respectively.

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The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2$ = 9.920 g

Mass of $H_2O$ = 2.031 g

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 9.920 g of carbon dioxide, $=\frac{12}{44}\times 9.920g=2.705g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 2.031 g of water, $=\frac{2}{18}\times 2.031g=0.226g$ of hydrogen will be contained.

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Molar mass of C = 12 g/mol

Molar mass of H = 1 g/mol

Putting values in equation 1, we get:

$\text{Moles of C}=\frac{2.705g}{12g/mol}=0.225 mol$

$\text{Moles of H}=\frac{0.226g}{1g/mol}=0.226 mol$

Step 2: Calculating the mole ratio of the given elements.

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.225 moles

$\text{Mole fraction of C}=\frac{0.224}{0.225}=1$

$\text{Mole fraction of H}=\frac{0.226}{0.225}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula of the compound becomes $K_1Mn_1O_4=KMnO_4$

To calculate the molecular formula, the number of atoms of the empirical formula is multiplied by a factor known as valency that is represented by the symbol, 'n'.

$n =\frac{\text{Molecular mass}}{\text{Empirical mass}}$ .....(2)

We are given:

Mass of molecular formula = 26.04 g/mol

Mass of empirical formula = 13 g/mol

Putting values in equation 3, we get:

$n=\frac{26.04g/mol}{13g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{1\times 2}H_{1\times 2}=C_2H_2$

Hence, the empirical and molecular formula for the given organic compound are $CH$ and $C_2H_2$ respectively.

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