Question

Report your answer with the proper number of significant figures. magnesium is used in lightweight alloys for airplane bodies and other structures. the metal is obtained from seawater in a process that includes precipitation, neutralization, evaporation, and electrolysis. how many kilograms of magnesium can be obtained from 8.01 km3 of seawater if the initial mg2+concentration is 0.12% by mass? (d of seawater = 1.04 g/ml)

Answer 1

First, convert 8.01 km^3 to L 


(1) 1 km^3 = (10^3 m)^3 = 10^9 m^3 


1 m^3 = 10^3 L 


10^9 m^3 = (10^9 m^3)*(10^3 L/m^3) = 10^12 L 


So 8.01 km^3 = 8.01 x 10^12 L 


(2) calculate the mass of that volume of seawater (1.04 g/mL = 1.04 kg/L) 

(8.01 x 10^12 L)*( 1.04 kg/L) = 8.3304 x 10^12 kg 

(3) calculate the Mg in that mass of water 

(8.3304 x 10^12 kg)*(0.0012) = 9.99648 x 10^10 kg of Mg