Question

A 0.250-g sample of a magnesium-aluminum alloy dissolves completely in an excess of HCl(aq). When the liberated H2(g) is collected over water at 29 ∘C and 752 torr, the volume is found to be 303 mL . The vapor pressure of water at 29 ∘C is 30.0 torr. How many moles of H2 can be produced from x grams of Mg in magnesium-aluminum alloy? The molar mass ofMg is 24.31 g/mol. Express your answer in terms of x to four decimal places (i.e., 0.5000x).

Answer 1

mass of Mg = x , mass of Al = y

x + y = 0.250 ------------------->1

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

24g                                                         2 g

x g                                                           a g

2 x = 24 a

a = 0.083 X g -------------------------->2

2 Al + 6 HCl ------------------------> 2 AlCl3 + 3H2

54g                                                                6 g

y g    b g

54 b = 6 y

b = 0.11 y --------------------------->3

moles of H2 :

T = 29 + 273 = 302 K

V = 303 mL = 0.303 L

P = 752-30 / 760 = 0.95 atm

P V = n R T

0.95 x 0.303 = n x 0.0821 x 302

n = 0.0116

moles of H2 = 0.0116

mass of H2 = 0.0232 g

0.083 x + 0.11y = 0.0232 g -------------------------> 4

    x + y = 0.250 ------------------->1

x = 0.159 g

y = 0.091 g

mass of Mg = 0.159 g

moles of Mg = 0.159 / 24 = 6.625 x 10^-3

a = 0.083 x

mass of H2 = 0.083 x 0.159 = 0.0132 g

moles of H2 = 6.599 x 10^-3

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

0.159 g Mg -----------------------> 6.599 x 10^-3 mol H2

x g mg -------------------------------

moles of H2 = 0.0415 x