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. a. 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) b. How many moles of gas are in the tube? c. Using the vapor pressure of H_2 O from Table 14.2, what is the pressure of H_2 in the eudiometer? d. How many moles of H_2 were produced? e. Assuming complete reaction with excess HCL, how many grams of Mg were used in the reaction?

Explanation / Answer

1. 29.96 mmHg = (1 atm * 29.96 mmHg) / 760 mmHg

= 0.04 atm

2. Given, P = 0.04 atm, V= 35.7ml = 0.0357 L, T= 21.5 C = 294.5 K , R = 0.0821 atm.L/mol.K

Using Ideal gas law, n = PV/RT

after solving n = 0.00143 / 24.18 /mol

= 0.00006 moles

3. Vapour pressure of H2O at given temperature = 18.7 torr = 0.0246 atm

(1 torr = 0.00131579 atm)

4.

Now, P = 0.0246 atm, V= 35.7ml = 0.0357 L, T= 21.5 C = 294.5 K , R = 0.0821 atm.L/mol.K

Using Ideal gas law, n = PV/RT

after solving n = 0.00088 / 24.18 /mol

= 0.0000363 moles

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