Oblicz ile kilogramow gliceryny ulegnie spaleniu jezeli powstanie 132 kg tlenku wegla (4) mc- 12u mh-1u mO- 16 u
2 C3H5(OH)3 + 7O2 -> 6CO2 + 8H2O
M CO2 = 44g = 0,044kg
6 * 0,044kg = 0,264kg
M C3H5(OH)3 = 92g = 0,092kg
2* 0,092kg = 0,184kg
x kg - 132kg
0,184kg - 0,264kg
0,264x = 24,288
x = 92kg
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2 C3H5(OH)3 + 7O2 -> 6CO2 + 8H2O
M CO2 = 44g = 0,044kg
6 * 0,044kg = 0,264kg
M C3H5(OH)3 = 92g = 0,092kg
2* 0,092kg = 0,184kg
x kg - 132kg
0,184kg - 0,264kg
0,264x = 24,288
x = 92kg