初三化学专题:初中化学计算题解题方法

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:1.2.1233.1231.=100=100:=100Vpm1w1=m2w2m1m2w1w21.MAB1=B+B2=M+2.1234().))2.12345312g=g3456=1-()105055.4xFe+H2SO4()=FeSO4+H2m()56256-2=5410x55.4g-50g=5.4gx=56%56%11210102COO2CO29ml1ml3.5mlCOO2CO23COCO23.44.4COCO2CO24COCO218CO2225AFeBAlCBaOH2DNa2CO3112.25Zn+H2SO4()ZnSO4+H22KClO3=2KCl+3O22H2+O2=2H2O2KClO33O26H26ZnKClO33ZnxKClO33Zn122.536512.25gxx==19.5g2__________________2Na+2H2O=2NaOH+H2Mg+2HCl=MgCl2+H22Al+6HCl=2AlCl3+3H23FeNaClFeFe2O3(FeFe)(1)(2)(1)8.0gH2SO41.2g(3)(2)NaOH8.0gFe2O3(FeFe2O3)(1)8.0gFeFe2O3(2)Fe(3)1Na2CO33.8%Na2O5.8%NaOH90.4%M18.75%10030%NaOH%29.252FeOH22004.9%15.23M4.0194520%50%M97.25%4M30%N[NM/48]5KOHH2O7.62%K2CO32.38%k2O10%KOH80%W9820%2010%KOH34.86MgOH220036.5%957143.54885CuR2520%R525%R88%1Fe2O350%AMgOBNa2OCCO2DSO226.80.4AAlMgBNaFeCZnCuDMgZn340%ANH4ClBCONH22CNH4HCO3DNH42SO445673%100AMgOZnOBCaOCuOCMgOCuODCaOMgO550mLPH=8PH3APH=0BPH=7CPH=14DPH=5,,,,,,,,,.,,32%A.20%B.40%C.48%D.80%1(NH4)2SO428%(A.CO(NH2)2B.NH4NO3C.NH4HCO3D.NH4Cl2C5H10C3H6OC6H12O630%A.60%B.45%C.30%D.22.5%3.55018.25%2.5()A.24B.40C.56D.655018.25%=9.1259.125H2=0.251H23.54=142.54=102820A1COCO2COCO2N224A10B30C50D702FeOFe2O3CaCO356CaCO3()A10B25C30D355R2O330%RA27B23C39D561R%=130%=70%R7323R=56DRXRX2R44.1%R34.5%AR3XB.RX3C.R2XD.RX3wNaHCO3NH4HCO3w/2NaHCO3NH4HCO32NaHCO3=Na2CO3+H2O+CO2NH4HCO3=NH3+H2O+CO2Na2CO31061062=212mNaHCO3=168mNH4HCO3=212-168=4410040%?(CuSO4*5H2O)CuSO4160250100%=64%(0%)x64%40%(100+x)100x=60RORCl2m554.0(9.5-4.0)m=4040-16=24CH4C2H487%A.CH4B.C2H2C.C2H6D.C8H8147.65%AKClO3BNaClCAlCl3DKCl50%25.6g0.18gA.B.C.D.KClO3KCl+O2236KClO32O232KClO3KCl+3O22KCl2C2H2+O2CO2+H2OO22H2O24C2H222C2H2+O2==CO2+2H2OCO24O252C2H2+5O24CO2+2H2OFe+H2OFe3O4+H2Fe3O4Fe3O4FeFeOH2OFe3H2O43Fe+4H2OFe3O4+H2H243Fe+4H2OFe3O4+4H21xyzxyzxyzFeS2+O2____Fe2O3+SO2FeS21,O2xFe2O3ySO2z=Fe:1=2yS:12=zO:2x=3y+2zx=11/4,y=1/2,z=24FeS2+11O2====2Fe2O3+8SO2

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