Question

Step 1. Obtain a 600 mL beaker, andadd 300 mL of water.

Step 2. Add 30 mL of HCl (2M) to the beaker, and stir.

Step 3. Add 10 mg of magnesium metal to the beaker.

Step 4. Allow thehydrogen gas to evolve. All of the magnesium should be consumed.

Step 5. Record the amount of hydrogen gas that has evolved using the chemical property dialog.

questions:

In your lab report, include the following information in the Discussion section of your report:

Record the moles of hydrogen gas that has evolved. Calculate the atomic weight of magnesium as follows:

• atomic weight of Mg (in grams) = weight (in grams) of Mg consumed / moles ofH₂ evolved

Discuss the stoichiometry of the chemical reaction that you are studying: Mg + 2HCl --> H2 + Mg2+ (aq) + 2Cl- (aq) How does the information in the balanced chemical equation above help you determine the atomic weight of magnesium? Discuss how the calculated atomic weight of the magnesium compare to information found on the periodic table of elements. <Explain, analyze and interpret what you observed. Draw conclusions from what you know. This is also where you show the reader that you understand the significance or meaning of the results. Refer to iLab instructions for items to include in your discussion.> < Within a few sentences, provide a concluding statement about the results of your laboratory>

Answer 1

Given reactions is

Mg + 2HCl --> H2 + Mg2+ (aq) + 2Cl-(aq)Hydrogen H2 (g) (0.060480g=0.030000moles

Magnesium Mg (solid) (9.270850g = 0.381438 moles

Magnesium Chloride MgCI2 (in solution)2.856300g=(0.030000moles)

Water H2O (liquid) (330g) =(18.318069 moles)

Moles of hydrogen evolved = 0.0300 moles

a) Atomic weight of Mg = weight of Mg consumed / moles of H2

= (10g - 9.27085 g ) / 0.0300 mol

=24.305 g/mol

b)

Number of molecules of H₂that were produced in the reaction

= moles of H₂* Avogadro's number

$\(=0.03mol+*+(6.023*10^{23})\)$

No. of H2 molecules= \(1.806 * 10^{22}\)

c)

Number of atoms of Magnesium that were involved in the reaction

= moles of Mg * Avagadro's number

$\(=(0.7291/23.305)*(6.023*10^{23})\)$ . atoms.

= \(1.8069 * 10^{23}\) atoms.

Answer 2

the reaction is given as:Mg + 2HCl --> H2 + Mg2+ (aq) + 2Cl-.now,we have the amount of H2 evolved:0.4166milimol,as the reacvtion suggests that 1mol Mg:1 mol H2,so,the atomic weight of Mg; (10/0.4166)= 24.0038 gm/mol