Gas law and moles

Question

A sample of solid magnesium reacts with excess $\ce{HCl}$ solution to form hydrogen gas. The sample of gas measures $1.628\ \mathrm{L}$ at $22.0\ \mathrm{^\circ C}$ and $0.930\ \mathrm{atm}$ of pressure.
a) how many moles of gas were collected?
b) how many grams of magnesium did you start out with?

First I stated by writing the chemical equation

$\ce{Mg + HCl -> H2 + MgCl}$

then convert the temperature to kelvin and pressure to atm and volume to L

$V = 1.628\ \mathrm{L}$
$T = 295.15\ \mathrm{K}$
$P = 0.930\ \mathrm{atm}$

$PV = nRT$

$0.930\ \mathrm{atm} \cdot 1.628\ \mathrm{l} = n \cdot 0.08206\ \mathrm{l\ atm\ K^{-1}\ mol^{-1}} \cdot 295.15\ \mathrm{K}$
which will give that $n = 16.0\ \mathrm{mol}$. Did I do this correctly?

Answer 1

Although it doesn’t affect the result, your equation is incorrect. Magnesium tends to form $\ce{Mg^2+}$ ions and chlorine tends to form $\ce{Cl-}$ ions. That’s why magnesium chloride compounds are found in 1:2 proportions. After balancing the coefficients your equation should look like this:

$\ce{Mg + 2HCl -> MgCl2 + H2}$

Then using the formula $pV = nRT$ we get $n\approx0.063\ \mathrm{mol}$

Since $n(\ce{Mg})=n(\ce{H2})$ we have $0.063\ \mathrm{mol}$ of magnesium, which is approximately $1.531\ \mathrm{g}$.

Answer 2

For (a)

$$n_{\ce{H2}} = \frac{p V}{R T} = \frac{(0.930\ \text{atm}) (1.628\ \text{L})}{R \times (22.0\ \text{K} + 273.15\ \text{K})} = \frac{(0.930\ \text{atm}) (1.628\ \text{L})}{R \times 293.2\ \text{K}} = 0.06251389252288155\ \text{mol}$$

With the correct number of significant figures

$n_{\ce{H2}} = 0.06251\ \text{mol}$.

For (b)

the balanced reaction tells you that 1 mole of $\ce{Mg}$ makes 1 mole of $\ce{H2}$, so $n_{\ce{Mg}} = 0.06251\ \text{mol}$. To get the mass of $\ce{Mg}$ use the atomic weight

$m_{\ce{Mg}} = n_{\ce{Mg}} \times M_{\ce{Mg}} = 0.06251\ \text{mol} \times M_{\ce{Mg}} = 1.519400157768636\ \text{g}$.

Again, don’t forget we only have four significant figures

$m_{\ce{Mg}} = 1.519$ g