Chemistry, 30.07.2019 02:00 jose477

How many liters of oxygen are needed to exactly react with 25.8 g of methane at stp? ch4(g) + 2 o2(g) â co2(g) + 2 h2o(l)?

Answers: 1

Show answers

Answers

The volume of oxygen ( ${{\text{O}}_2}$ ) that react with 25.8 g of methane at STP is $\boxed{{\text{72}}{\text{.059 L}}}$ .

Further explanation:

Balanced chemical reaction:

The chemical reaction that contains equal number of atoms of the different elements in the reactant as well as in the product side is known as balanced chemical reaction. The chemical equation is required to be balanced to follow the Law of the conservation of mass.

Stoichiometry of a reaction is used to determine the amount of species present in the reaction by the relationship between the reactants and products. It can be used to determine the moles of a chemical species when the moles of other chemical species present in the reaction is given.

Consider the general reaction,

${\text{A}}+2{\text{B}}\to3{\text{C}}$

Here,

A and B are reactants.

C is the product.

Mole is the S.I. unit and number of moles is calculated as the ratio of mass of the compound to that of molar mass of the compound. Molar mass is the mass of different atoms present in a molecule.

The expression of number of moles, mass and molar mass of compound is as follows:

${\text{Number of moles}}=\frac{{{\text{mass of the compound}}}}{{{\text{molar mass of the compound}}}}$ …... (1)

At standard temperature and pressure, 1 mole of gas produced in a balanced chemical reaction occupies 22.4 liters of volume.

Given mass of ${\text{C}}{{\text{H}}_4}$ is 25.8 g.

Molar mass of ${\text{C}}{{\text{H}}_4}$ is 16.04 g/mol.

Substitute these values in equation (1) to calculate the number of moles of ${\text{C}}{{\text{H}}_4}$ .

$\begin{aligned}{\text{number of moles}}&=\frac{{{\text{25}}{\text{.8 g}}}}{{{\text{16}}{\text{.04 g/mol}}}}\\&={\text{1}}{\text{.608478 mol}}\\\end{aligned}$

The balanced chemical reaction between ${\text{C}}{{\text{H}}_4}$ and ${{\text{O}}_2}$ is as follows:

${\text{C}}{{\text{H}}_4}\left(g\right)+{\text{2}}{{\text{O}}_2}\left(g\right)\to{\text{C}}{{\text{O}}_2}\left(g\right)+{\text{2}}{{\text{H}}_2}{\text{O}}\left(l\right)$

According to stoichiometry of the above balanced chemical reaction, 1 mole of methane ( ${\text{C}}{{\text{H}}_4}$ ) react with 2 moles of oxygen ( ${{\text{O}}_2}$ ) to produce 1 mole of carbon dioxide ( ${\text{C}}{{\text{O}}_2}$ ) and 2 moles of water ( ${{\text{H}}_2}{\text{O}}$ ).

Therefore, ratio of mole of methane ( ${\text{C}}{{\text{H}}_4}$ ) to oxygen ( ${{\text{O}}_2}$ ) will be 1:2.

The formula to calculate number of moles of oxygen ( ${{\text{O}}_2}$ ) is as follows:

${\text{Number of moles }}\left({{{\text{O}}_2}} \right)=2\times{\text{number of moles of C}}{{\text{H}}_4}$ …… (2)

Substitute1.608478 mol for number of moles of methane ( ${\text{C}}{{\text{H}}_4}$ ) in equation (2).

$\begin{aligned}{\text{Number of moles}}\left( {{{\text{O}}_2}} \right)&=2\times{\text{1}}{\text{.608478 mol}}\\&={\text{3}}{\text{.21695}}\;{\text{mol}}\\\end{aligned}$

At STP, one mol of gas occupies 22.4 liters of volume.

Number of moles of oxygen ( ${{\text{O}}_2}$ ) is 3.21695 mol.

The volume of 3.21695 mol of oxygen ( ${{\text{O}}_2}$ ) is calculated as follows:

$\begin{aligned}{\text{Volume}}&=\left({\frac{{{\text{22}}{\text{.4}}\;{\text{L}}}}{{1\;{\text{mol}}}}}\right)\left({{\text{Volume of one mole of }}{{\text{O}}_2}}\right)\\&=\left({\frac{{{\text{22}}{\text{.4}}\;{\text{L}}}}{{1\;{\text{mol}}}}}\right)\left( {{\text{3}}{\text{.21695}}\;{\text{mol}}}\right)\\&={\text{72}}{\text{.05968}}\;{\text{L}}\\&\approx{\text{72}}{\text{.059}}\\\end{gathered}$