Дано
m(пракC6H14)=344 g
η=90%
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m(Na)-?
m(C3H7CL)-?
m(теор)=m(прак)*100% / 90=344*100%/90=382.22 g
X Y 382.22
2C3H7CL+2Na-->C6H14 +2NaCL
2*78.5 2*23 86.5
M(C3H7CL)=78.5 g/mol
M(Na)=23 g/mol
M(C6H14)=86 g/mol
X=2*78.5*382.22 / 86.5
X=693.74 g - C3H7CL
Y=2*23*382.22 / 86.5
Y=203.26g - Na
ответ 693.74 г - C3H7CL , 203.26 г - Na