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?

2 Answers
Nov 28, 2015

Answer:

#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?

Nov 28, 2015

Answer:

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.