Question

Gas experiement: Mg + HCl reaction

The volume of hydrgeon gas produced was not at standard temperautre and pressure. Convert the volume at the pressure and temperature of the experiement to a volume at standard temperature and pressure : STP (T=273.15 and P=1.00 atm).

Mg + 2HCl -> MgCl2 + H2

Mass of Mg: .012

Max pressure: 1.0946

Initial pressure: 1.0103

Pressure change: .0843

Temperature: 294.3

Used 5 ml of 1.0 M HCl in experiment

Answer 1

Given reaction is

Given mass of Mg = 0.012 g

Pressure = 1.0946

Initial pressure = 1.0103

Pressure change = 0.0843

T = 294.3

Volume of HCl = 5 mL , [HCl]= 1.0 M

Calculation of limiting reactant

Mol of Mg = 0.012 g x 1 mol / 24.31 g

=0.0004994 mol

Mol of HCl = volume in L x molarity

= 0.005 L x 1.0 M = 0.005 mol HCl

Lets show chemical reaction

Mg + 2 HCl (aq) -- > MgCl2 + H2

Moles of HCl for 0.0004994 mol Mg

= 0.0004994 mol Mg x 2 mol HCl / 1 mol Mg

= 0.0004994 mol HCl

But in actual there is 0.005 mol HCl so Mg is limiting reactant

Moles of H2 formation depends on Mg

Lets calculate moles of H2

Moles of H2 = moles of Mg x 1 mol H2 / 1 mol Mg

Moles of H2 = 0.0004994 mol

Now we get volume of H2

pV = nRT

V = nRT / p

= [0.0004994 x 0.08206 x 294.3 /1.0946] L

= 0.0110 L

So the volume of H2 = 11.0 mL

Now we convert this volume to volume to STP

P1V1 T2 = P2 V2 T1

V2 = 1.0946 x 0.0110 x 273.15 /(1 x 294.3)

=0.0112 L

Volume of H2 at STP = 0.0112 L