Question

Aluminum reacts with excess hydrochloric acid to form aqueous aluminum chloride and 190.6 ml of hydrogen gas (collected over water at 26°C and 0.89 atm). (Vapor pressure of water at 26ºC = 25.2 mmHg.) How many grams of aluminum reacted? Enter to 4 decimal places.

Answer 1

Answer:

0.1296 grams of aluminum reacted

Explanation:

Equation of reaction:

2Al + 6HCl = 2AlCl3 + 3H2

From the ideal gas equation

PV = nRT

n = PV/RT

P = (25.2/760)atm + 0.89atm = 0.923 atm

V = 190.6 ml = 190.6cm^3

R = 82.057cm^3.atm/gmol.K

T = 26°C = 26+273K = 299K

n = 0.923×190.6/82.057×299 = 0.0072 moles of H2

From the equation of reaction

3 moles of hydrogen gas is formed from 2 moles of aluminum

Therefore, 0.0072 moles of hydrogen gas is formed from (0.0072×2/3 = 0.0048) moles of aluminum

Number of moles of aluminum that reacted = 0.0048 moles

Mass of aluminum that reacted = number of moles × MW = 0.0048 × 27 = 0.1296 grams

Answer 2

Answer:0.119g

Explanation:equation of rxn is

2Al+6HCl=2AlCl3+3H2

From ideal gas eqn

PV=nRT

n=PV/RT

P here is the partial pressure of H2 from the qtn.According to Dalton law of partial pressure, PT=PH2+PH20

PT=0.89atm given

PH20=25.2mmhg given=25.2/760atm,=0.033atm

PH2=PT-PH20

PH2=0.89-0.033=0.857atm

T=26+273=299K

R=0.082atmdm^-3mol^-1K^-1

V=190.6ml=190.6cm3=190.6/1000=0.1906dm3

n=PV/RT

n=0.857*0.1906/0.082*299

=0.00667moles of H2.

From the eqn of reaction,

2moles of Al reacts to gv 3moles of H2

xmoles of Al will give 0.00667moles of H2

xmoles=0.00667*2/3 (cross multiplying)=0.00444moles of Al

From the relationship, n=mass/MW

mass=MW*n

MW of Al=27g/mol

mass=0.0044moles*27g/moles

mass=0.119grams of Al.