What volume of hydrogen gas is produced when "6 g" of magnesium reacts?

Aug 8, 2017

Well, I got a more exact answer... but you have only allowed yourself one significant figure. Sorry to say, I can only give your answer to zero decimal places.

${V}_{{H}_{2} \left(g\right)} \approx \text{6 L}$...

I assume you mean ${25}^{\circ} \text{C}$ and $\text{1 atm}$, but you know your room temperature better than anyone around you. As for the reaction, I can only assume what it is... Is it not this one?

${\text{Mg"(s) + 2"HCl"(aq) -> "MgCl"_2(aq) + "H}}_{2} \left(g\right)$

If we are on the same page, then for every $\text{1 mol}$ of $\text{Mg}$ consumed, $\text{1 mol}$ of ${\text{H}}_{2} \left(g\right)$ is made. This was found from the stoichiometry of the reaction, i.e. $\textcolor{red}{1} \times {\text{Mg"(s) harr color(red)1 xx "H}}_{2} \left(g\right)$.

This means that from the molar mass of $\text{Mg}$:

6 cancel"g Mg" xx cancel"1 mol Mg"/(24.305 cancel"g Mg") xx ("1 mol H"_2)/(cancel"1 mol Mg")

$= {\text{0.2467 mols H}}_{2}$ were produced.

Now, at this point, we would assume that ${\text{H}}_{2} \left(g\right)$ is an ideal gas at ${25}^{\circ} \text{C}$ and $\text{1 atm}$, and use the ideal gas law:

$P V = n R T$

where:

• $P$ is the pressure in $\text{atm}$.
• $V$ is the volume in $\text{L}$.
• $n$ is the $\boldsymbol{\text{mols}}$ of ideal gas.
• $R = \text{0.082057 L"cdot"atm/mol"cdot"K}$ is the universal gas constant with the appropriate units.
• $T$ is the temperature in $\text{K}$.

And thus,

${V}_{{H}_{2} \left(g\right)} = \frac{n R T}{P}$

= (("0.2467 mols H"_2)("0.082057 L"cdot"atm/mol"cdot"K")(25 + "273.15 K"))/("1 atm")

$= {\text{6.0396 L H}}_{2}$

Unfortunately, you have only allowed yourself a measly one significant figure. :( I'm sorry.

You can only say you have $\textcolor{b l u e}{{\text{6 L H}}_{2}}$...