Answer to Question #114592 in General Chemistry for Litzy

Question

Show calculations for each step.

0.080 mg of Mg reacts with HCl (aq) to give 85.5 mL of og H2 gas at 22oC and 759 mm Hg pressure.

Mg(s) + 2HCl (aq) LaTeX: \longrightarrow
⟶
MgCl2 (aq) + H2(g)

a) What is the partial pressure of H2 gas?

b) Use CGL to calculate volume of H2 gas at STP.

(hint: make a table with correct units for lab conditions and STP conditions and then solve for V2).

c) Calculate moles of Mg and get moles of H2 gas using mol-mol ratio from the balanced equation.

d) Calculate molar volume of H2 gas (Hint: divide V2 by moles of H2).

e) Calculate % error (theoretical value = 22.4 L/mol)

Answer 1

a)

From step (c): n (H2) = (Mg) = 0.00329 mol

PV = nRT

P1 = 759 mm Hg = 101,191.678 Pa

T1 = 22 + 273.15 = 295.15 K

V = nRT / P

V = 0.00329 * 8.314 * 293.15 / 101,191.678

V = 79.78 mL - vlume of H2 in the mixture

79.78 / 85.5 * 101,191.678 Pa = 94.4 kPa - partial pressure of H2

b)

STP: 273.15 K and 101,325 Pa

CGL: P1V1/T1 = P2V2/T2

V2 = P1V1T2/P2V2

P1 = 759 mm Hg = 101,191.678 Pa

V2 = (101,191.678*79.78*(22+273.15))/(101,325*273.15)

V2 = 86.1 mL

c)

M (Mg) = 24.31 g/mol

n = 0.080 / 24.31 = 0.00329 mol

d)

V2 = 86.1 mL

Vm = 86.1 mL / 0.00329 mol = 26.2 (L/mol)

e)

Theoretical value = 22.4 L/mol

% error = (26.2 - 22.4)/22.4 = 0.17 or 17%