a 1.78 gram sample of an unknown gas is found to occupy a volume of .945 L at a

Question

a 1.78 gram sample of an unknown gas is found to occupy a volume of .945 L at a pressure of 645 mmhg and a temperature of 33 Celsius. Assume ideal Behavior the molecular weight of the unknown gas is?

in units of u

part B:
sodium metal reacts with water to produce hydrogen gas the product gas H2 is collected over water at a temperature of 20 Celsius and a pressure of 759 mmhg. if the wet H2 gas formed occupies a volume of 9.29 L the number of grams of H2 formed is ______g. the water vapor pressure is 17.5 mmhg at 20 Celsius.

part C:
sodium metal reacts with water to produce hydrogen gas the product gas H2 is collected over water at a temperature of 25 Celsius and a pressure of 742 mmhg if the wet H2 gas formed occupies a volume of 7.14 L the number of grams of H2 formed is______g. the vapor pressure of water is 23.8 mmhg at 25 Celsius.

Explanation / Answer

a]

PV = nRT

P in atm

V in L

T in kelvin ; R = 0.0821

n = moles = mass / molarmass

1 atm = 760 mm Hg

[645/760]*0.945 = 1.78*0.0821*(33+273) / Molar mass

Molar mass = 55.758 gm /mol

b]

2Na + 2H2O ----> 2NaOH + H2

Ptotal = 759 mm Hg

Ptotal = P­H2 + PH2O                    T = 20.0 °C = 293 K

pH2 = 759 - 17.5 = 741.5 mm Hg

In atm ;

741.5 / 760 = 0.9756 atm

PV = nRT

0.9756 * 9.29 = n*0.0821*(293)

n = 0.37677

1 mole of H2 = 2 gms

0.37677 moles of H2 = 0.75354 gms

C]

pH2 = 742 - 23.8 = 718.2 mm Hg

pH2 in atm = 718.2 / 760 = 0.945 atm

PV = nRT

0.945*7.14 = n*0.0821*(25+273)

n = 0.275

1 mole of H2 = 2 gms

0.275 moles of H2 = 0.551 gms of H2 is formed

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.