Question

Magnesium (Mg) is a valuable metal used in alloys, in batteries, and in the manufacture of chemicals. It is obtained mostly from seawater, which contains about 1.3 g of Mg for every kilogram of seawater. Referring to Problem 1.87, calculate the volume of seawater (in liters) needed to extract 8.0 x 10^4 tons of Mg, which is roughly the annual production in the United States.

Answer 1

6.0 × 10¹⁰ L

Step-by-step explanation:

We know that 1 ton = 10⁶ g. 8.0 × 10⁴ tons of Mg are:

8.0 × 10⁴ ton × (10⁶ g/ 1 ton) = 8.0 × 10¹⁰ g

There is about 1.3 g of Mg for every kilogram of seawater. The mass of seawater that contains 8.0 × 10¹⁰ grams of Mg is:

8.0 × 10¹⁰ g Mg × (1 kg seawater/ 1.3 g Mg) = 6.2 × 10¹⁰ kg seawater

The average density of seawater is 1.025 kg/L. The volume corresponding to 6.2 × 10¹⁰ kg is:

6.2 × 10¹⁰ kg × (1 L/ 1.025 kg) = 6.0 × 10¹⁰ L

Answer 2

the volume of sea water is 85866.407 L, necessary to extract 8.0 E4 tons of Mg

Step-by-step explanation:

• %p/p = 1.3 g Mg / 1000g seawater ) * 100 = 0.13 %

⇒ 8.0 E4 ton Mg * ( 0.00110231 Kg / ton ) = 88.1848 Kg Mg

⇒ 0.13 % = ( 88.1848 Kg Mg / Kg seawater ) * 100

⇒ ( 0.13 / 100 ) = 88.1848 / Kg seawater

⇒ Kg seawater = 88.1848 / 1 E-3

⇒ Kg seawater = 88184.8 Kg

⇒ Vseawater = mass seawater / density seawater

∴ density seawater = 1027 Kg / m³...from literature

⇒ V seawater = 88184.8 Kg / 1027 Kg/m³

⇒ V seawater = 85.866 m³ * ( 1000L / m³ )

⇒ V seawater = 85866.407 L