2C2H2+O2--->4C+2H2O d=m/v m=d*v=10*1,07=10,7g MC2H2=24+2=26G/MOL=2*26=52g 52g---32g 10,7--x x=6,58g do spalenia etynu potrzeba 6,58g O2 10,7g----y 52g------48g y=9,87g powstanie 9,87g wegla
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2C2H2+O2--->4C+2H2O
d=m/v
m=d*v=10*1,07=10,7g
MC2H2=24+2=26G/MOL=2*26=52g
52g---32g
10,7--x
x=6,58g
do spalenia etynu potrzeba 6,58g O2
10,7g----y
52g------48g
y=9,87g
powstanie 9,87g wegla