Question

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This for u.. ❤

sulphide of Cobalt metal has a cubic structure with four formula units per unit cell. if density is 4.26 g/cc and edge length of unit cell is 6.93 A. determine mass of Sulphur required to produce 1.5 kg of this compound

Pls ans it✌

Answer 1

4.26 × 1000 = 4M ( 1.67 ×10^-27)/ (6.93× 10^-10)^3

42600 = 4 M × 1.67 × 10^ -27/ 332.8 × 10^-30

42600 = 6.67 × 10^-27 M/ 332.8× 10^-30

14177280 × 10-30/ 6.67 × 10^-27 = M

212552.2 × 10^ -3

212.6

M= 59 x + 32 y = 212.6

x= 1 ,y = 2 M = 59 + 64 = 113

x= 2, y= 3 M = 118 + 96 = 214

approximately same so take x= 2, y= 3

Co2 S3

it gives 3 moles S

So moles of S = 3× 1.5/ 214 × 10^3

= 4500/ 214 = 21.02

mass= 32 × 21.02 =672.64 g

Answer 2

mass=672.64g

hope it helps you