Question

Oblicz stosunki wagowe pierwiastków w związkach chemicznych:
Fe2S3-
K2O-
Oblicz zawartość % pierwiastków w związku chemicznym:
N2O3-
Ca(OH)2-
SO3-
Fe2S3-

Answer 1

Fe₂S₃ - 2*56u:3*32u=112:96=7:6
K₂O - 2*39u:16u=78:16=39:8

mN₂O₃=2*14u+3*16u=76u
%N=28u*100%/76=36,84%
%O=48u*100%/76=63,16%

mCa(OH)₂=40u+2*16u+2u=74u
%Ca=40u*100%/74=54,05%
%O=32u*100%/74=43,25%
%H=2u*100%/74=2,7%

mSO₃=32u+3*16u=80u
%S=32u*100%/80=40%
%O=48u*100%/80=60%

mFe₂S₃=2*56u+3*32u=208u
%Fe=112u*100%/208=53,85%
%S=96u*100%/208=46,15%