*- do sześcianu
Oblicz ile gramów i ile dm* w dwutlenku węgla (CO2) powstanie, jeżeli spaliny w tlenie 20g węgla.
C + O₂ -> CO₂
M(C) = 12
1 mol węgla --- 1 mol dwutlenku węgla
1,66 mola węgla --- 1,66 mola dwutlenku węgla
M(CO₂)=12 g/mol + 16g/mol*2=12g/mol+32g/mol=44g/mol
m=1,66mola*44g/mol=73,33g
1 mol --- 22,4dm³
1,66 mola ---x
x = 22,4dm³*1,66 = 37,33dm³
M(C)=12g
M(CO2)=12g+2*16g=44g
C+O2-->CO2
12g------------44g
20g------------xg
xg=20g*44g/12g
xg=73,3g
12g--------------22,4dm³
20g----------------xdm³
xdm³=20g*22,4dm³/12g
xdm³=37,3dm³
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C + O₂ -> CO₂
M(C) = 12
1 mol węgla --- 1 mol dwutlenku węgla
1,66 mola węgla --- 1,66 mola dwutlenku węgla
M(CO₂)=12 g/mol + 16g/mol*2=12g/mol+32g/mol=44g/mol
m=1,66mola*44g/mol=73,33g
1 mol --- 22,4dm³
1,66 mola ---x
x = 22,4dm³*1,66 = 37,33dm³
M(C)=12g
M(CO2)=12g+2*16g=44g
C+O2-->CO2
12g------------44g
20g------------xg
xg=20g*44g/12g
xg=73,3g
12g--------------22,4dm³
20g----------------xdm³
xdm³=20g*22,4dm³/12g
xdm³=37,3dm³