Question

Calculati nr de atomi de carbon care se gasesc in 680 g de piatra de var cu 20% impuritati.

Answer 1

Teorie: puritate=m pura/m impura*100
Piatra de var este carbonatul de calciu, CaCO3
1 mol de atomi conttine 6.022*10 la 23 atomi 
Calcule:
puritate=m pura/m impura*100=>m pura=680*80/100=544 g CaCO3 pur
n=m/masa molara=544/100=5.44 moli CaCO3
1 mol carbonat contine 6.022*10 la 23 atomi carbon
5.44 moli carbonat contine x atomi carbon
x=32.76*10 la 23 atomi carbon
Raspuns:
In 680 g piatra de var, de puritate 80% se gasesc 32.76*10 la 23 atomi carbon

Answer 2

p=mpur/mimpur*100==>mpur=p*mimpur/100=80*680/100=544g CaCO3

MCaCO3=ACa+ACa+3AO=40+12+3*16=100g/mol

n=m/M=544/100=5,44 moli CaCO3
 
1mol C.........................NA=6.023*10²³ atomi C
5,44moli C....................x=32.76512*10²³ atomi C