Question

methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. what volume of hydrogen chloride would be produced by this reaction if 7.0 ml of methane were consumed? also, be sure your answer has a unit symbol, and is rounded to 2 significant digits.

Explanation:

Step1: Write the balanced chemical equation

The reaction between methane ($\ce{CH4}$) and chlorine ($\ce{Cl2}$) to form carbon tetrachloride ($\ce{CCl4}$) and hydrogen chloride ($\ce{HCl}$) is:
$$\ce{CH4(g) + 4Cl2(g) -> CCl4(g) + 4HCl(g)}$$
From the equation, 1 mole (or volume, for gases at constant T and P) of $\ce{CH4}$ produces 4 moles (or volumes) of $\ce{HCl}$.

Step2: Relate volumes of $\ce{CH4}$ and $\ce{HCl}$

At constant temperature and pressure, the volume ratio of gases in a reaction is equal to their mole ratio (Avogadro’s law). So, if volume of $\ce{CH4}$ is $V_{\ce{CH4}} = 7.0\ \text{mL}$, the volume of $\ce{HCl}$ ($V_{\ce{HCl}}$) is:
$$V_{\ce{HCl}} = 4 \times V_{\ce{CH4}}$$

Step3: Calculate $V_{\ce{HCl}}$

Substitute $V_{\ce{CH4}} = 7.0\ \text{mL}$:
$$V_{\ce{HCl}} = 4 \times 7.0\ \text{mL} = 28\ \text{mL}$$
(2 significant digits, as 7.0 has 2)

Answer:

$28\ \text{mL}$