Question

69 kg Na reactioneaza cu Cl
a) Numarul de moli de color utilizati
b) Masa de sare formata de puritate 80%
c)Procentul de clor din sarea formata masic

Answer 1

a) 2Na+Cl2=2NaCl

m Na=69kg

A( masa atomică) Na = 23 Kg/kmol

N kmoli = A/m

N kmoli =69/23=3 kmoli

A Cl =35,5

1 kmol Cl2 =2 * 35,5 = 71 Kg

2 kmol Na......1 kmol Cl2

3 kmoli Na......x kmoli Cl2

X=1,5 kmoli Cl2 (1kmol=1000moli)

1,5 kmoli =1500moli Cl2

b) M NaCl = 23+35,5=58,5

2moli NaCl=2 * 58,5 = 117 kg

23 Kg Na.........117 Kg NaCl

69.....................y

y = 351 kg

351×80/100= 280.8

c) 117 Kg NaCl .....71Kg Cl

100Kg NaCl........z

z = 60,68 % Cl