Question

5. Compozitia Procentuala 1. CaCO3; 2. CH4;​

Answer 1

Răspuns:

CaCo3

masa moleculara:

M CaCo3 = A Ca + A C + 3 × A O

=40+12+3×16

=40+12+48

=100 g

compozitia procentuala:

in 100g CaCo3...40gCa...12gC...48gO

x=100×40 supra 100=40%Ca

y=100×12 supra 100=12%C

z=100×48 supra 100=48%O

CH4

masa moleculara:

M CH4 = A C + 4 × A H

= 12 + 4 × 1

= 12 + 4

= 16

compozitia procentuala:

daca in 16 g CH4...12gC...4gH

atunci in 100 g CH4...xgC...ygH

x=100×12 supra 16=75gC

y=100×4 supra 16=25gH