Zadanie 1
a) mCH4=12u+4*1u=16u
Stosunek masowy:
mC:mH=12:4=3:1
b) mN2O5=2*14u+5*16u
mN:mO=28:80=7:20
c) mFe2O3=2*56u+3*16u
mFe:mO=112:48=7:3
Zadanie 2
a) mNH3=14u+3*1u=17u
14u---x%
17u---100%
x=82,35%N
100%-82,35%=17,65%H
b) mK2O=2*39u+16u=94u
78u---x%
94u---100%
x=82,98%K
100%-82,98%=17,02%O
c) mCaO=40u+16u=56u
40u---x%
56u---100%
x=71,43%Ca
100%-71,43%=28,57%O
