Oblicz objętości tlenku węgla (IV) i amoniaku (warunki normalne) potrzebnych do otrzymania 1 tony mocznika.
M(CO(NH₂)₂)=12g+16g+2*14g+4*1g=60g=0,06kg
1t=1000kg
CO₂+2NH₃-->CO(NH₂)₂+H₂O
22,4dm³ CO₂ ---------- 0,06kg CO(NH₂)₂
xdm³ CO₂ --------------- 1000kg CO(NH₂)₂
x=22,4dm³*1000kg/0,06kg
x=373333,3dm³=373,3m³ CO₂
2*22,4dm³ NH₃ ---------- 0,06kg CO(NH₂)₂
xdm³ NH₃ ------------------ 1000kg CO(NH₂)₂
x=44,8dm³*1000kg/0,06kg
x=733333,3dm³=733,3m³ NH₃
mCO(NH2)2=60u
CO2 + 2NH3 ---> CO(NH2)2 + H2O
22,4dm3 CO2 ------- 60g CO(NH2)2
xm3 CO2 ------------ 1000kg CO(NH2)2
x = 373,3m3 CO2
44,8dm3 NH3 ------- 60g CO(NH2)2
xm3 NH3 ---------- 1000kg CO(NH2)2
x = 733,3m³ NH3
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M(CO(NH₂)₂)=12g+16g+2*14g+4*1g=60g=0,06kg
1t=1000kg
CO₂+2NH₃-->CO(NH₂)₂+H₂O
22,4dm³ CO₂ ---------- 0,06kg CO(NH₂)₂
xdm³ CO₂ --------------- 1000kg CO(NH₂)₂
x=22,4dm³*1000kg/0,06kg
x=373333,3dm³=373,3m³ CO₂
2*22,4dm³ NH₃ ---------- 0,06kg CO(NH₂)₂
xdm³ NH₃ ------------------ 1000kg CO(NH₂)₂
x=44,8dm³*1000kg/0,06kg
x=733333,3dm³=733,3m³ NH₃
mCO(NH2)2=60u
CO2 + 2NH3 ---> CO(NH2)2 + H2O
22,4dm3 CO2 ------- 60g CO(NH2)2
xm3 CO2 ------------ 1000kg CO(NH2)2
x = 373,3m3 CO2
44,8dm3 NH3 ------- 60g CO(NH2)2
xm3 NH3 ---------- 1000kg CO(NH2)2
x = 733,3m³ NH3