Question

С  водным раствором гидроксида натрия взаимодействует каждое из двух веществ:1)Ba(OH)2 и CO22)Al(OH)2 и HCl3)Fe(Oh)2 и CaO4)Zn(OH)2 и MgCO3К окислительно-восстановительным  относится реакция1)H2+S=H2S2)SO2+H2O=H2SO33)Cu(OH)2+H2SO4=CuSO4+2H2O4)AgNO3+NaCl=AgCl+NaNO3масса осадка,образующегося при взаимодействии избытка раствора сульфата алюминия с раствором,содержащим  4,16 г хлорида бария,равна

Answer 1

1 --------  2) АІ(ОН)3 і НСІ
2 --------  1)
3 ---------   4,66 г

Al2(SO4)3 + 3BaCL2 = 2ALCL3 + 3BaSO4
                     3*208  г --------------- 3*233 г
                       4,16 г -----------------   х г
х = 4,16*699/624 = 4,66 г

Answer 2

1) Ba(OH)2 + CO2 -> BaCO3 + 2H2ODouble displacement reaction

2) Al(OH)2 + 2HCl -> AlCl2 + 2H2ODouble displacement reaction

3) Fe(OH)2 + CaO -> FeO + Ca(OH)2 Double displacement reaction

4) Zn(OH)2 + MgCO3 -> ZnCO3 + Mg(OH)2Double displacement reaction

The oxidizing-reducing reaction is: 2) SO2 + H2O -> H2SO3 + Reduction of SO2

To find the mass of the precipitate formed when excess aluminum sulfate solution is added to 4.16 g of barium chloride solution:

The balanced equation for this reaction is:

Al2(SO4)3 + 3BaCl2 -> 2AlCl3 + 3BaSO4

The molar mass of BaSO4 is 233.38 g/mol (1 Ba, 1 S, and 4 O). The molar mass of BaCl2 is 208.23 g/mol (1 Ba and 2 Cl).

First, calculate the number of moles of BaCl2 in 4.16 g:

4.16 g / 208.23 g/mol = 0.02 mol BaCl2

Since the reaction ratio of BaCl2 to BaSO4 is 3:3, the number of moles of BaSO4 formed is also 0.02 mol.

Now, using the balanced equation, we can see that 3 moles of BaSO4 are formed for every 1 mole of Al2(SO4)3 used. Therefore, the number of moles of Al2(SO4)3 needed is:

0.02 mol / 3 = 0.0067 mol Al2(SO4)3

The molar mass of Al2(SO4)3 is 342.15 g/mol (2 Al, 3 S, and 12 O). The mass of Al2(SO4)3 needed is:

0.0067 mol x 342.15 g/mol = 2.32 g Al2(SO4)3

Therefore, when excess aluminum sulfate is added to 4.16 g of barium chloride, 2.32 g of BaSO4 precipitate will be formed.