Chemistry, 31.01.2020 08:52 kenken2583

What mass of ccl4 is formed by the reaction of 5.14 g of methane with an excess of chlorine?

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Because CH4 is the limiting reagent, we must find how many moles we have.
1. 5.14 g CH4 * (1 mol CH4 / 16.04 g/mol) = .32125 moles CH4
Now we know that there are .32125 moles of CCl4 produced as well.
Next, take .32125 moles and multiply it by the atomic mass of CCl4, which is 153.82 grams = 49.29 g CCl4 produced by the reaction.

Answer from: sierravick123owr441

$\boxed{{\text{49}}{\text{.224 g}}}$ of ${\text{CC}}{{\text{l}}_4}$ is produced by the reaction of 5.14 g of methane with an excess of chlorine.

Further explanation:

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.

One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

Limiting reagent:

A limiting reagent is the one that is completely consumed in a chemical reaction. The amount of product formed in any chemical reaction has to be in accordance with the limiting reagent of the reaction. The amount of product depends on the amount of limiting reagent since the product formation is not possible in the absence of it.

The given reaction occurs as follows:

${\text{C}}{{\text{H}}_4}\left(g\right) + 4{\text{C}}{{\text{l}}_2}\left(g\right)\to{\text{CC}}{{\text{l}}_4}\left(g\right)+4{\text{HCl}}\left(g\right)$

It is given that chlorine is present in an excess amount so methane $\left({{\text{C}}{{\text{H}}_4}}\right)$ is a limiting reagent and therefore it controls the amount of ${\text{CC}}{{\text{l}}_4}$ produced during the reaction.

The formula to calculate the amount of ${\text{C}}{{\text{H}}_4}$ is as follows:

${\text{Amount of C}}{{\text{H}}_4} = \frac{{{\text{Given mass of C}}{{\text{H}}_{\text{4}}}}}{{{\text{Molar mass of C}}{{\text{H}}_{\text{4}}}}}$ …… (1)

The given mass of ${\text{C}}{{\text{H}}_4}$ is 5.14 g.

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

Substitute these values in equation (1).

$\begin{aligned}{\text{Amount of C}}{{\text{H}}_4} &= \left({{\text{5}}{\text{.14 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{16}}{\text{.04 g}}}}}\right)\\ &= 0.32044\\&\approx{\mathbf{0}}{\mathbf{.320 mol}}\\\end{aligned}$

From the balanced chemical reaction, it is clear that 1 mole of ${\text{C}}{{\text{H}}_4}$ produces 1 mole of ${\text{CC}}{{\text{l}}_4}$ .

So the amount of ${\text{CC}}{{\text{l}}_4}$ is equal to that of ${\text{C}}{{\text{H}}_4}$ and thus the amount of ${\text{CC}}{{\text{l}}_4}$ is 0.320 mol.

The formula to calculate the mass of ${\text{CC}}{{\text{l}}_4}$ is as follows:

${\text{Mass of CC}}{{\text{l}}_4} = \left( {{\text{Moles of CC}}{{\text{l}}_4}}\right)\left({{\text{Molar mass of CC}}{{\text{l}}_4}}\right)$

…… (2)

The number of moles of ${\text{CC}}{{\text{l}}_4}$ is 0.320 mol.

The molar mass of ${\text{CC}}{{\text{l}}_4}$ is 153.82 g/mol.

Substitute these values in equation (2).

$\begin{gathered}{\text{Mass of CC}}{{\text{l}}_4} = \left({{\text{0}}{\text{.320 mol}}}\right)\left({\frac{{{\text{153}}{\text{.82 g}}}}{{{\text{1 mol}}}}}\right)\\ = {\mathbf{49}}{\mathbf{.224 g}}\\\end{gathered}$

Therefore, the mass of ${\text{CC}}{{\text{l}}_4}$ produced is 49.224 g.

Learn more:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride:

2. How many grams of potassium were in the fertilizer?

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: stoichiometry, reactant, product, limiting reagent, CCl4, CH4, Cl2, molar mass, given mass, amount of CH4, amount of CCl4, 49.224 g, mass of CCl4, 4Cl2, 4 HCl, 5.14 g, 16.04 g/mol.

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