Chemistry magnesium reaction with HCI

Question

0.0367g of magnesium to produce hydrogen gas collected over water at 24°C and 746mmHg atmospheric pressure. volume of the collected gas was 46.3mL, what is the value of the gas constant, R?
Percent yield = 100%
What is the % error(when compared with the true value, R=0.0821 L-atm/K-mol)?

Answer 1

Mg + 2HCl ==> MgCl2 + H2(g)

moles Mg = 0.0367 g x 1 mole/24.3 g = 1.51x10-³ moles

moles H2 produced = 1.51x10-3 moles Mg x 1 mole H2/mole Mg = 1.51x10^-3 moles H2 (assuming HCl is not limiting)

PV = nRT and solve for R

P = 746 mm Hg - 22.4 mm Hg = 723.6 mm Hg (subtracted vapor pressure of water at 24ºC) = 723.6/760 = 0.952 atm

V = 46.3 ml x 1 L/1000 ml = 0.0463 L

n = 1.51x10^-3 moles 

T = 24 + 273 = 297K

R = PV/nT = (0.952 atm)(0.0463 L)/(1.51x10^-3 mole)(297K)

R = 0.0983 L-atm/K-mol

% error = (0.0983 - 0.0821)/0.0821 = 0.1973 x100% = 19.7% error