Part 1 Synthesis of MgCl 2 : Weigh 0.1000g of Mg is mixed with 6ml of 3M HCl Wha

Question

Part 1 Synthesis of MgCl2:

Weigh 0.1000g of Mg is mixed with 6ml of 3M HCl

What is the total molecular reaction for the reaction of Mg and HCl (don't know what they mean by that :-/) ???

Part 2 Synthesis of Mg(OH)2:

Magnesium chloride can be converted to magnesium hydroxide by adding sodium hydroxide.

Pour the MgCl2 solution made in part 1 into large test tube. Measure 3.5ml of 6M NaOH into a clean 10ml graduated cylinder. To isolate the Mg(OH)2 a centrifuge will be used. Be sure to save Mg(OH)2 to the next step.

What is the total molecular equation for the reaction of MgCl2 and NaOH???

What will be the theoretical yield of Mg(OH)2???

Part 3 Synthesis of MgSO4:

Magnesium sulfate can be synthesized from magnesium hydroxide by adding sulfuric acid. Add 5.0ml of 2.0 M H2SO4 to a clean 10ml graduated cylinder and then pour it into the test tube containing the solid Mg(OH)2.

What will be the total molecular equation for the reaction of Mg(OH)2 and H2SO4???

What will be the theoretical yield of MgSO4???

Part 4 Synthesis of MgCO3:

Magnesium carbonate can be synthesized from magnesium sulfate by adding sodium carbonate.

Pour the MgSO4 mixture from part 3 into a clean 50 ml Erlenmeyer flask.To the flask add 20ml of 1.0 M Na2CO3 solution.

What will be the total molecular equation for the reaction of MgSO4 and NaCO3???

Will be really grateful for help cuz am really bad at equations :((((((((

It would be nice to see all the calculations for the rest so will be able to solve it later on my own :))))

Explanation / Answer

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

2) MgCl2 (aq) + 2 NaOH (aq) = Mg(OH)2 (s) + 2 NaCl (aq)

Atomic masses: Mg = 24.31, Cl = 35.45, Na = 22.99, O = 16.00, H = 1.008, S = 32

So, MW of Mg(OH)2 = (24.31 + 2*(16+1.008)) = 58.326

Theoretical yield is the amount of product if reaction goes to completion (100%).

Here 1 mole of Mg is producing 1 mole of Mg(OH)2

So, 24.31 g Mg is producing 58.326 g Mg(OH)2

Thus, 0.1 g Mg will produce 58.326 x 0.1 / 24.31 g = 0.2399 g Mg(OH)2

3) Mg(OH)2(s) + H2SO4(aq) = MgSO4(aq) + 2 H2O(l)

Calculate theoretcial yield similarly.

MW of MgSO4 = 24.31 + 32 + 4*16 = 120.31

1 mole of Mg is producing 1 mole of MgSO4

24.31 g of Mg gives 120.31 g MgSO4

So, 0.1 g Mg gives 120.31 x 0.1 / 24.31 g = 0.4949 g of MgSO4

4) MgSO4(aq) + Na2CO3(aq) = MgCO3(s) + Na2SO4(aq)

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