Based on the data provided and the reaction Mgs 2HClaq MgC

Based on the data provided and the reaction:

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

Mg (g) = 0.0324 g

HCl (4.00 M) = 30.0 ml

Temp (degree C) = 23.0

V invert (ml) = 21.32

Atmosphereic pressure (kPa) = 101.8

P h2o (mmHg) = 21.07

V gas (ml) = 27.82

R= 0.08206 L atm mol^-1 k^-1

1 atm = 760 mm

Hg = 1.01325x10^5 Pa

MW H2 = 2.02 g/mol

*Calculate the theoretic yield of H2 gas.

*Calculate the actual yield.

*Calculate the percent yield.

theoratical yield ----- 1 mole of Mg yields 1 moleof H2 so .0027 gr of H2 forms

practical yield----------- PV=NRT p= 1 atm v=.0021 L R=R= 0.08206 L atm mol^-1 k^-1 T=273+23K

so .0017 g H2 released

% of yield = practicl yield/theoritical yield---------(0 .0017/.0027)*100=63 %