Question

Aluminum reacts with excess hydrochloric acid to form aqueous aluminum chloride and hydrogen gas. The resulting gas is collected over water and it measured to have 218.6 ml at 26°C and 0.947 atm. If the vapor pressure of water is 25.2 mmHg at 26ºC, how many grams of aluminum reacted?

Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules. Several other gases make up the remaining 1% of air molecules.

What is the partial pressure of nitrogen in air at an atmospheric pressure of 0.986 atm? Assume ideal behavior.

Express your answer numerically, to three significant figures.

please sdhow work so i can attempt the next set of problems

Answer 1

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

PV = nRT

n   = W/M

T   = 26C⁰     = 26 + 273    = 299K

V   = 0.2186L

vapor pressure of water at 26C⁰   = 25.2mmHg   = 25.2/760   = 0.033atm

Atmospheric pressure                  = 0.947atm

pressure of H2                            = Atmospheric pressure-vapor pressure of water at 26C⁰

                                                     = 0.947-0.033   = 0.914atm

M                                                = 2

PV = WRT/M

W    = PVM/RT

        = 0.947*0.2186/0.0821*299    = 0.0084g of H2

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

3 moles of H2 produced from 2 moles of Al

3*2g of H2 produced from 2*27g of Al

0.0084g of H2 produced from = 2*27*0.0084/3*2    = 0.0756g of Al >>>>>Answer

2.

atmospheric pressure of 0.986 atm

Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules

The pressure of 78% Of N2 = 78/100*0.986    = 0.76908atm of N2