Question

CH4(g) +4Cl2(g) ....> CCl4(g) + 4HCl(g) What mass of CCl4 is formed by the reaction of 5.14 g of methane with an excess of chlorine?

Answer 1

There are a few steps to this problem.

1.First you convert the reactant's mass to moles.

2.Next you use a ratio of the moles and stoichiometric coefficients of the species of interest.

Solve the unknown moles in the ratio (this would be moles of CCl4 for your problem)

3. Convert the moles of CCl4 to mass (grams).

Answer 2

Let me elaborate on 2.

Set up a ratio using the species of interest, like so:

e.g. for a general reaction:

\(\sf \color{red}{a}A + \color{blue}{b}B\) \(\rightleftharpoons\) \( \color{green}{c}C\)

where upper case are the species (A,B,C), and lower case (a,b,c) are the coefficients ,

\(\sf \dfrac{n_A}{\color{red}{a}}=\dfrac{n_B}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\) (n=moles)

From here you can isolate what you need.
For example: if you have 2 moles of B, how many moles of C can you produce?

solve algebraically: \( \sf\dfrac{2}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\rightarrow n_C=\dfrac{2*\color{green}{c}}{\color{blue}{b}}\)