The total pressure in the eudiometer was equilibrated with the lab pressure, the

Question

The total pressure in the eudiometer was equilibrated with the lab pressure, the temperature was 21.5 degree C, and the volume measured was 35.7 mL. The pressure in the lab was assumed to be the same as the weather report of 29.96 in Hg. What is that pressure in atm? (760 mmHg = 1 atm) How many moles of gas are in the tube? Using the vapor pressure of H_2O from Table 14.2, what is the pressure of H_2 in the eudiometer? How many moles of H_2 were produced? Assuming complete reaction with excess HCL, how many grams of Mg were used in the reaction?

Explanation / Answer

Answer (a)

According to weather report , pressure in lab is 29.96 inch Hg = 760.984 mmHg = 1.001 atm

Answer (b)

The number of moles of hydrogen gas collected can then be calculated from the ideal gas law:

n = PV /RT

Here

P = P total=PH2+ PH2O =1.001 atm

V = 0.0357 L

R = 0.0821 L atm/mol K

T = (21.5 + 273 )K = 294.5 K

so n = (1.0246 atm * 0.0357 L) / (0.0821 L atm/mol K * (294.5)K

   n = 0.00151 moles of gas are in the tube

Answer (C)

Determine pressure of dry H2(PH2) by subtract ing the vapor pressure of water from the total pressure

P total=PH2+ PH2O =PH2+ 0.0246 atm =1.001atm

PH2 = 0.9764 atm

Answer (D)

n = PV /RT

Here

P = PH2= 0.9764 atm

V = 0.0357 L

R = 0.0821 L atm/mol K

T = (21.5 + 273 )K = 294.5 K

so n = (0.9764 atm * 0.0357 L) / (0.0821 L atm/mol K * (294.5)K

   n = 0.00144 moles of H2 were produced

Answer (E)

Convert # moles Mg reacted to moles of H2 that could be produced. (1 mole H2is produced for every1 mole Mg reacted-this comes from the balanced chemical equation)

Mg (s) + 2HCl (aq)MgCl2(aq) + H2(g)

so n = 0.00144 moles of H2 were produced = 0.00144 moles of Mg were used in the reaction

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