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mC2H5NH2=45u
C2H5Cl + NH3 ---> C2H5NH2 + HCl
1mol C2H5Cl ------ 45g C2H5NH2
xmoli C2H5Cl ------ 200g C2H5NH2
x = 4,44mola C2H5Cl
NH3 + C2H5Cl --> C2H5NH2 + HCl
x 200g
z proporocji wyliczam x
x = 4,44 mola