A student carried out this experiment using 0.0795 g of pure magnesium ribbon. A

Question

A student carried out this experiment using 0.0795 g of pure magnesium ribbon. At the end of the reaction 81.4 mL of H_2 was collected at 22 degree C and the barometric pressure was 755.0 torr. The student also measured the height (h) of liquid column in audiometer (above the level in beaker) to be 5.0 cm. (Vapor pressure of water at 22 degree C = 19.8 torr, R = 0.08206^1.08/mol K) (a) Calculate the partial pressure of H_2 gas collected. (b) Calculate the volume of H_2 gas collected under STP condition. (c) Calculate the molar volume of H_2 gas at STP form this reaction.

Explanation / Answer

2) The experiment aims to show the reactivity of magnesium toward dilute HCl. Mg reacts with dil. HCl to evolve hydrogen gas (H2). The reaction taking place is

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

Copper metal is less reactive than hydrogen gas and hence will not evolve hydrogen from HCl. However, iron is more reactive than hydrogen and will certainly evolve hydrogen gas from HCl. Therefore, if Fe wire is used, both Mg and Fe will react to produce H2. The reaction of iron with HCl takes place as

2 Fe (s) + 6 HCl (aq) ----> 2 FeCl3 (aq) + 3 H2 (g)

3) Write down the balanced chemical reaction:

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

As per the balanced stoichiometric equation,

1 mole Mg = 2 moles HCl.

Molar mass of Mg = 24 g/mol.

Moles of Mg taken in the experiment = (0.0795 g)/(24 g/mol) = 0.0033125 mole.

Therefore, moles of HCl required for complete reaction = (0.0033125 mole Mg)*(2 moles HCl/1 mole Mg) = 0.006625 mole.

Therefore, volume of 6.0 M HCl required = moles of HCl required/concentration of HCl = (0.006625 mole)/(6.0 mol/L) = 0.001104 L = (0.001104 L)*(1000 mL/1 L) = 1.104 mL 1.10 mL (ans).

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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