Question

Magnesium (Mg) is a valuable metal used in alloys, in batteries, and in chemical synthesis. It is obtained mostly from seawater, which contains about 1.3 g of Mg for every kilogram of seawater. Calculate the volume of seawater (in liters) needed to extract 8.0 x 104 tons of Mg, which is roughly the annual production in the United States. (rho seawater = 1.03 g/ml).

Answer 1

Answer:

The volume of seawater needed to extract $8.0\times 10^4$ tons of magnesium is $5.420\times 10^{10} L$.

Explanation:

Concentration magnesium in sea water = 1.3 g /kg of seawater

Extracted mass of magnesium ,= $8.0\times 10^4 tons$

(1 ton = 907185 g)

$8.0\times 10^4 tons=8.0\times 10^4\times 907185 g$

$=7.257\times 10^{10} g$

If 1 kg of sea water contains 1.3 grams of magnesium. Then $7.257\times 10^{10} g$ of magnesium will be contained by:

$\frac{1}{1.3} kg\times 7.257\times 10^{10}=5.582\times 10^{10} kg$ sea water.

Mass of sea water,m =$5.582\times 10^{10} kg=5.582\times 10^{13} g$

Volume of sea water = v

Density of sea water ,d= 1.03 g/mL

$v=\frac{m}{d}=\frac{5.582\times 10^{13} g}{1.03 g/mL}=5.420\times 10^{13} mL$

1 mL = 0.001 L

$v = 5.420\times 10^{10} L$

The volume of seawater needed to extract $8.0\times 10^4$ tons of magnesium is $5.420\times 10^{10} L$.