need help answering questions from gas constant experiment. please show your wor

Question

need help answering questions from gas constant experiment. please show your work for #3 and explain the others thoroughly

QUESTIONS FOR GAS CONSTANT EXP NAME There are several sources of error in this experiment that are unavoidable What are these errors and what can be done to keep them as small as possible? 1· What must be done if the amount of Mg is miscalculated and is so large that the gas gen the tube? 2. tube is larger than the 786.8 mmHg. The level 3. A student used o.o938 g of Mg and collected 96.21 mL of H2 gas (their eudiometer (their eu one we used) over water at 29.3°C on a day when the atmospheric pressure was of the solution in the tube was 12.6 mm above the level in the beaker when the reaction was e experimental value for R in units of L mmHg moli- K-1? How does this compare (calculate percent error) to the accepted value for R?

Explanation / Answer

Write the balanced chemical equation for the reaction.

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

As per the stoichiometric equation,

1 mole Mg = 1 mole H2.

Atomic mass of Mg = 24.305 g/mol; molar mass of H2 = 2*1.008 g/mol = 2.016 g/mol.

Mole(s) of Mg corresponding to 0.0938 g Mg = (0.0938 g)/(24.305 g/mol) = 0.0038593 mole.

Mole(s) of H2 = (0.0038593 mole Mg)*(1 mole H2/1 mole Mg) = 0.0038593 mole.

We need to ascertain the pressure of dry hydrogen gas which is given by the equation

P = Pa – PH2O - Pleveldifference

The atmospheric pressure, Pa is 786.8 mmHg, PH2O = 30.0 mmHg; the level difference is 12.6 mm; this can be converted to pressure difference as

Plevel difference = (12.6/13.5) mmHg = 0.93 mmHg; therefore, P = (786.8 – 30.0 – 0.93) mmHg = 755.87 mmHg.

Use the ideal gas equation to find out the value of the gas constant, R.

P*V = n*R*T where P = 755.87 mmHg; V = 96.21 mL = (96.21 mL)*(1 L/1000 mL) = 0.09621 L; T = 29.3°C = (29.3 + 273.15) K = 302.45 K and n = 0.0038593 mole.

(755.87 mmHg)*(0.09621 L) = (0.0038593 mole)*R*(302.45 K)

====> R = 62.30246 L.mmHg/mol.K (ans).

The accepted value of R in L.mmHg/mol.K is 62.36367 L.mmHg/mol.K.

Percent error = (accepted value) – (experimental value)/(accepted value)*100 [… denotes the absolute value without any sign]

= (62.36367 L.mmHg/mol.K) – (62.30246 L.mmHg/mol.K)/(62.36367 L.mmHg/mol.K)*100

= 0.098% (ans).

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

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