Reaction of HCl (aq) and Mg(s) to form MgCl2(aq) and H2(g) Given: Molarity of HC

Question

Reaction of HCl (aq) and Mg(s) to form MgCl2(aq) and H2(g)

Given:

Molarity of HCL : 3.00M

volume of HCL: 20.0mL

mass of Mg: 0.036g

volume of gas before placing in equalization chamber: 37.0mL

volume of gas after placing in equalization chamber: 37.5 mL

barometric pressure of the room: 736.4 mmhg

temperature of the room: 18.5 C

------------------------------------------------

Calculate:

mole of Mg reacted:

mole of H2(g) formed:

vapor pressure of vater:

pressure of H2 (Daltons Law):

pressure of H2 in atm:

volume of H2 in L:

Temperature in K:

Calculation of R:

Percent Error of R:

Explanation / Answer

Molarity of HCl = 3 M

Volume of HCl = 20 mL

Moles of HCl = 3 * 20 = 60 mmoles

Mass of Mg = 0.036 g

Moles of Mg = 0.036 / 24.3 * 1000 = 1.48 mmoles

2 HCl + Mg -> MgCl2 + H2

By stoichiometry,

a)

Moles of Mg reacted = 1.48 mmoles

b)

Mole of H2 formed = 1.48 mmoles

c)

Volume occupied by H2 = 37.5 mL = 0.0375 mL

Pressure of room, P = 736.4 mmHg = 736.4 / 760 atm = 0.969 atm

Pressure of H2 = P = 736.4 mmHg

d)

Pressure of H2 in atm = 736.4 / 760 = 0.969 atm

e)

Volume of H2 in L = 37.5 / 1000 = 0.0375 L

f)

Temperature = 18.5 C

= 18.5 + 273.15 K = 291.65 K

g)

P * V = n * R * T

0.969 * 0.0375 = 1.48 x 10-3 * R * 291.65

R = 0.084 L atm/mol-K

h)

True value of R = 0.082 L atm/mol-K

% error in R = (0.084 – 0.082 / 0.082) * 100

= 2.4 %

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