challenging gas law problem?

Question

a sample of magniesium metal was allowed to react with excess hydrochloric acid solution. the gas liberated was collected over water at 23degrees celceius. The volume of gas collected was 47.1 mL and the atmospheric pressure in the lab was 744 mm Hg. the vapor pressure of the water at 23 degrees is 21.1 mm Hg.

a) write a ballanced equation for the reaction described above.

b) what volume in mL would the gas occupy if it was dryed and stored at STP

c)determinethe mass of magneisium metal in grams that would need to be reacted to liberate the quantity of the gas collected above

tell the equations used in your calculations.

Answer 1

a) Mg + 2HCl --> MgCl2 + H2

b) this is a PVT or combined gas law problem
P1 = 744 - 21.1 = 722.9 mm
V1 = 47.1 mL
T1 = 23C = 296 K
P2 = 760 mm
V2 = ?
T2 = 273 K

V2 = V1 x (P1/P2) x (T2/T1)
V2 = 47.1 mL x (722.9 mm/760 mm) x (273 K/296 K)
V2 = 41.3 mL at STP

c)
convert volume to L
41.3 mL = 0.0413 L H2

convert L H2 to mole H2
0.0413 L H2 x (1 mole H2 / 22.4 L H2) = 0.00184 mole H2

convert mole H2 to mole Mg
0.00184 mole H2 x (1 mole Mg / 1 mole H2) = 0.00184 mole Mg

convert mole Mg to grams Mg
0.00184 mole Mg x (24.3 g Mg / 1mole Mg) = .0447 g Mg

Answer 2

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

b) 744mmHg - 21.1mmHg = 723mmHg

23C = 296K

47.1mL x 723mmHg/760mmHg x 296K/273K = 48.6mL

c. 48.6mLH2 x 1molH2/22,400mLH2 x 1molMg/1molH2 x 24gMg/1molMg = 0.052g Mg