Ile moli tlenu potrzeba do całkowitego spalenia 7,8 g benzenu ? C - 12 u , H - 1u . O - 16 u .
C6H6 + O2 --> CO2 + H20
7,8g - xmol
78g/mol - 1mol
x= 0,1mol
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C6H6 + O2 --> CO2 + H20
7,8g - xmol
78g/mol - 1mol
x= 0,1mol