# In the equation Mg(s) + 2HCl(aq) -> MgCL_2(aq) + H_2(g), what mass of hydrogen will be obtained if 100 cm^3 of 2.00 mol dm^-3 HCl are added to 4.86 g of magnesium?

Nov 28, 2015

#### Answer:

$M g \left(s\right) + 2 H C l \left(a q\right) \rightarrow M g C {l}_{2} \left(a q\right) + {H}_{2} \left(g\right) \uparrow$

#### Explanation:

Moles of metal, $=$ $\frac{4.86 \cdot g}{24.305 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.200$ $m o l$.

Moles of $H C l$ $=$ $100 \cdot c {m}^{-} 3 \times 2.00 \cdot m o l \cdot {\mathrm{dm}}^{-} 3$ $=$ $0.200$ $m o l$

Clearly, the acid is in deficiency ; i.e. it is the limiting reagent, because the equation above specifies that that 2 equiv of HCl are required for each equiv of metal.

So if $0.200$ $m o l$ acid react, then (by the stoichiometry), 1/2 this quantity, i.e. $0.100$ $m o l$ of dihydrogen will evolve.

So, $0.100$ $m o l$ dihydrogen are evolved; this has a mass of $0.100 \cdot m o l \times 2.00 \cdot g \cdot m o {l}^{-} 1$ $=$ ??g.

If 1 mol dihydrogen gas occupies $24.5$ ${\mathrm{dm}}^{3}$ at room temperature and pressure, what will be the VOLUME of gas evolved?

Nov 28, 2015

#### Answer:

The limiting reactant is $\text{HCl}$, which will produce $\text{0.202 g H"_2}$ under the stated conditions.

#### Explanation:

This is a limiting reactant problem.

$\text{Mg(s)" + "2HCl(aq)}$$\rightarrow$$\text{MgCl"_2("aq")"+ H"_2("g")}$

Determine Moles of Magnesium
Divide the given mass of magnesium by its molar mass (atomic weight on periodic table in g/mol).

$4.86 \cancel{\text{g Mg"xx(1"mol Mg")/(24.3050cancel"g Mg")="0.200 mol Mg}}$

Determine Moles of 2M Hydrochloric Acid
Convert $\text{100 cm"^3}$ to $\text{100 mL}$ and then to $\text{0.1 L}$.
${\text{1 dm}}^{3}$$=$$\text{1 L}$
Convert ${\text{2.00 mol/dm}}^{3}$ to $\text{2.00 mol/L}$
Multiply $0.1 \text{L}$ times $\text{2.00 mol/L}$.

$100 \cancel{\text{cm"^3xx(1cancel"mL")/(1cancel"cm"^3)xx(1"L")/(1000cancel"mL")="0.1 L HCl}}$

$\text{2.00 mol/dm"^3}$$=$$\text{2.00 mol/L}$

$0.1 \cancel{\text{L"xx(2.00"mol")/(1cancel"L")="0.200 mol HCl}}$

Multiply the moles of each reactant times the appropriate mole ratio from the balanced equation. Then multiply times the molar mass of hydrogen gas, $\text{2.01588 g/mol}$

$0.200 \text{mol Mg"xx(1"mol H"_2)/(1"mol Mg")xx(2.01588"g H"_2)/(1"mol H"_2)="0.403 g H"_2}$

$0.200 \text{mol HCl"xx(1"mol H"_2)/(2"mol HCl")xx(2.01588"g H"_2)/(1"mol H"_2)="0.202 g H"_2}$

The limiting reactant is $\text{HCl}$, which will produce $\text{0.202 g H"_2}$ under the stated conditions.