The surface of the oxidized sodium is wrapped in aluminum foil, the small hole is stabbed, and the chemical formula is put into the water
(1) by oxidation of sodium and remove the oxide film with acupuncture holes in aluminum foil wrapped into the water, sodium hydroxide and water reacts with sodium hydroxide, sodium hydroxide and water reacts with hydrogen, aluminum and sodium hydroxide solution reaction of sodium metaaluminate and hydrogen, the reaction equation is: 2Na+2H2O=2NaOH+H2 = Na2O+H2O=2NaOH, 2Al+2NaOH+2H2O=2NaAlO2+3H2 = a:,, the formula is: Na2O+H2O=2NaOH, 2Na+2H2O=2NaOH+H2 = 2Al+2NaOH+2H2O=2NaAlO2+3H2, =;
(2) the reaction: Na2O+H2O=2NaOH, 2Na+2H2O=2NaOH+H2 = 2Al+, 2NaOH+2H2O=2NaAlO2+3H2 = n (NaOH), =2.0 L * 0.05 mol L-1=0.1? Mol, n (H2) =1.12L22.4L/mol=0.05 mol n (NaAlO2) =n (Al) =0.27mol27g/mol=0.01 mol, by electronic conservation knowledge: n (Na) +3n (Al) =2n (H2): n (Na) +3 * 0.01 mol=2 * 0.05 mol solution: n (Na) =0.07 mol by sodium conservation know: n (NaAlO2) +n (NaOH) =n (Na) +2n (Na2O): 0.01 mol+0.1 mol=0.07 mol+2n (Na2O), n (Na2O): =0.02 mol, w (Na) = (0.07mol+2 * 0.02mol) * 23g/mol0.07mol * 23g/mol+0.02mol * 62g/mol * 100% = 88.8%, a: the mass fraction of sodium block for sodium 88.8%.
(1) by oxidation of sodium and remove the oxide film with acupuncture holes in aluminum foil wrapped into the water, sodium hydroxide and water reacts with sodium hydroxide, sodium hydroxide and water reacts with hydrogen, aluminum and sodium hydroxide solution reaction of sodium metaaluminate and hydrogen, the reaction equation is: 2Na+2H2O=2NaOH+H2 = Na2O+H2O=2NaOH, 2Al+2NaOH+2H2O=2NaAlO2+3H2 = a:,, the formula is: Na2O+H2O=2NaOH, 2Na+2H2O=2NaOH+H2 = 2Al+2NaOH+2H2O=2NaAlO2+3H2, =; (2) the reaction: Na2O+H2O=2NaOH, 2Na+2H2O=2NaOH+H2 = 2Al+, 2NaOH+2H2O=2NaAlO2+3H2 = n (NaOH), =2.0 L * 0.05 mol L-1=0.1? Mol, n (H2) =1.12L22.4L/mol=0.05 mol n (NaAlO2) =n (Al) =0.27mol27g/mol=0.01 mol, by electronic conservation knowledge: n (Na) +3n (Al) =2n (H2): n (Na) +3 * 0.01 mol=2 * 0.05 mol solution: n (Na) =0.07 mol by sodium conservation know: n (NaAlO2) +n (NaOH) =n (Na) +2n (Na2O): 0.01 mol+0.1 mol=0.07 mol+2n (Na2O), n (Na2O): =0.02 mol, w (Na) = (0.07mol+2 * 0.02mol) * 23g/mol0.07mol * 23g/mol+0.02mol * 62g/mol * 100% = 88.8%, a: the mass fraction of sodium block for sodium 88.8%.
