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nFe=m/56 mol nFe=nFe(NO3)3 = m/56 mol mFe(NO3)3=242m/56 g nNO2=19.72/22.4=0.88 mol mNO2=40.48g Fe(NO3)3 = Fe3+ 3NO3- m/56-------------------->3m/56 nHNO3=nNO3- + nNO2=3m/56+0.88 suy ra mHNO3=(3m/56+0.88)*63 g nH2O=nHNO3/2=(3m/56+0.88)/2 suy ra mH2O=(3m/56+0.88)*9 g Bảo toàn khối lượng mhh + mHNO3 = mFe(NO3)3 + mNO2 + mH2O 49.6 + (3m/56+0.88)*63 = 242m/56 + 40.48 + 9(3m/56+0.88) suy ra m=39.648 g nFe=0.708 mol mFe(NO3)3=0.708*242=171.336 g
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