áp dụng ĐLBTKL => mH2 = 5,6+7,3-12,7=0,2(g)

\(\frac{13}{15}-\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{7}{10}\)

\(\frac{13}{15}-\left(\frac{13}{21}.\frac{7}{12}+\frac{7}{12}.x\right)=\frac{7}{10}\)

\(\frac{13}{15}-\left(\frac{13}{36}+\frac{7}{12}.x\right)=\frac{7}{10}\)

\(\frac{13}{15}-\frac{13}{36}-\frac{7}{12}.x=\frac{7}{10}\)

\(-\frac{7}{12}.x=\frac{7}{10}-\frac{13}{15}+\frac{13}{36}\)

\(-\frac{7}{12}.x=\frac{126}{180}-\frac{156}{180}+\frac{65}{180}\)

\(-\frac{7}{12}.x=\frac{7}{36}\)

\(x=\frac{7}{36}:\left(-\frac{7}{12}\right)\)

\(x=\frac{7}{36}.\frac{-12}{7}\)

\(x=\frac{-1}{3}\)

4FeS2 + 11O2 => 2Fe2O3 + 8SO2

a. nCu = 1 mol

nN = 1,2 = 2mol

nO = 3.2 = 6 mol

b.mCu = 1.64 = 64g

mN = 2.14 = 28 g

mO= 6.16 = 96 g

c. nCu(NO3) 2 = \(\frac{37,6}{188}=0,2\)

=> mCu = 0,2.64 = 12,8 (g)

mN = 0,2.2.14 = 5,6 (g)

nO = 0,2.6.16 = 19,2(g)

K2O + H2SO4=> K2SO4 + H2o

CuO + H2SO4=> CuSO4 +H2O

Al2O3 +3H2SO4 => Al2(SO4)3 + 3H2O

nSO4 =n H2SO4 = \(\frac{850.19,6\%}{98}=1,7mol\)

theo PTHH ta thấy n_{O(trong oxit)} = nSO4 = 1,7

=> m kim loại = m oxit - m_{O(trong oxit)}= 83,6 - 1,7.16 = 56,4(g)

=>m= m muối = mkim loại + mSO4 = 56,4 + 1,7.96 = 219,6(g)

nN2 = 0,03

m muối - mkim loại = mNO3

=> mNO3 = 54,9-7,5 = 47,4(g)

=> n_{N(trong muối)} = nNO3 = \(\frac{47,4}{62}=0,764mol\)

BT nguyên tố N => nHNO3 = n_{N(trong muối)} +n_N2 = 0,764 + 0,03.2 = 0,824 mol

=> V = 0,824 (l)

a. gọi CT AlxOy

ta có x:y = \(\frac{52,94\%}{27}:\frac{47,06\%}{16}=2:3\)

=> Al2O3

b. %O = 56,14%

gọi CT AlxSyOz

ta có x:y:z = \(\frac{15,79\%}{27}:\frac{28,07\%}{32}:\frac{56,14\%}{16}=2:3:12\)

=> Al2(SO4)3

nAl = 1/27 nCl2 = 1/71

2Al + 3Cl2 => 2AlCl3

=> Al dư

nAlCl3 = 2/3nCl2 = \(\frac{2}{3}.\frac{1}{71}\)= 2/213

=> mAlCl3 = \(\frac{2}{213}.133,5=\frac{89}{71}\left(g\right)\)

nNaOH = 0,375 mol

2NaOH + Ba(HCO3)2 => BaCO3 + Na2CO3 + 2H2O

0,375--------------------------> 0,1875---> 0,1875

a. m tủa = 0,1875.197 = 36,9375

b. m muối sinh ra = 0,1875.106 = 19,875