Question

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?

Answer 1

`Mg(s) + 2HCl(aq) rarr MgCl_2(aq) + H_2(g)uarr`

Explanation:

Moles of metal, `=` `(4.86*g)/(24.305*g*mol^-1)` `=` `0.200` `mol`.

Moles of `HCl` `=` `100 *cm^-3xx2.00*mol*dm^-3` `=` `0.200` `mol`

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` `mol` acid react, then (by the stoichiometry), 1/2 this quantity, i.e. `0.100` `mol` of dihydrogen will evolve.

So, `0.100` `mol` dihydrogen are evolved; this has a mass of `0.100*molxx2.00*g*mol^-1` `=` `??g`.

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

Answer 2

The limiting reactant is `"HCl"`, which will produce `"0.202 g H"_2"` under the stated conditions.

Explanation:

This is a limiting reactant problem.

`"Mg(s)" + "2HCl(aq)"` `rarr` `"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).

`4.86cancel"g Mg"xx(1"mol Mg")/(24.3050cancel"g Mg")="0.200 mol Mg"`

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

`100cancel"cm"^3xx(1cancel"mL")/(1cancel"cm"^3)xx(1"L")/(1000cancel"mL")="0.1 L HCl"`

`"2.00 mol/dm"^3"` `=` `"2.00 mol/L"`

`0.1cancel"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, `"2.01588 g/mol"`

`0.200"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"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 `"HCl"`, which will produce `"0.202 g H"_2"` under the stated conditions.