Question

Magnesium is obtained from sea water. Ca(OH)2 is added to sea water to precipitate Mg(OH)2. The precipitate is filtered and reacted with HCl to produce MgCl2. The MgCl2 is electrolyzed to produce Mg and Cl2. If 185 g of magnesium are recovered from 962 g of MgCl2, what is the percent yield for this reaction? Answer in units of percent

Answer 1

Answer is: the percent yield for this reaction is 75,37%.
m(MgCl₂) = 962 g.
n(MgCl₂) = m(MgCl₂) ÷ M(MgCl₂).
n(MgCl₂) = 962 g ÷ 95,21 g/mol.
m(MgCl₂) = 10,1 mol.
n(MgCl₂) : n(Mg) = 1 : 1.
n(Mg) = 10,1 mol.
m(Mg) = 10,1 mol · 24,3 g/mol.
m(Mg) = 245,43 g.
ω(Mg) = 185 g ÷ 245,43 g · 100% = 75,37%.

Answer 2

Answer:

The percentage yield of magnessium is 19.231 %

Explanation:

The magnesium from sea water is added to Ca(OH)2 to precipitate Mg(OH)2. This magnesium hydroxide is then reacted with HCl as follows ;

Mg(OH)2 + 2HCl → MgCl2 + 2H2O

molar mass of MgCl2 = 24.305 + 35.453 × 2 =  24. 305 + 70.906 = 95.211 g

The equation is balanced now . And 1 mole or 95.11 g of MgCl2 was produced . The magnesium chloride was electrolyzed   to produce Mg and Cl2.

The total mass of MgCl2 obtained = 962 grams

mass of the Mg obtained  = 185 grams

The percentage yield for the magnesium  = 185 / 962 × 100 = 18500/962

The percentage yield for the magnesium  = 185 / 962 × 100 = 19.231 %

The percentage yield of magnessium is 19.231 %