Ecuația reacției de ardere:
1.58 x
C + O₂ ------> CO₂
1 1
Rezolvare:
[tex]m_{C}=20\ g\ (m_{impur})\\M_{C}=12\ g/mol\\5\%\ impur.\implies\ P=95\%\\\\P=\frac{m_{pur}}{m_{impur}}\cdot100\implies m_{pur}=\frac{P\cdot m_{impur}}{100}=\frac{95*20}{100}=19\ g\ C\\\\n=\frac{m(pur)}{M}=\frac{19}{12}=1.58\ moli\ C\\\\n_{CO_2}=x=1.58\ moli\\M_{CO_2}=12+(2\cdot16)=12+32=44\ g/mol\\n=\frac{m}{M}\implies\ m=n\cdot M=1.58\cdot44=69.52\ g\ CO_2[/tex]
