Hỗn hơp X gam Na va Al4...

- Hướng dẫn giải

Al4C3+12H2O--->3CH4+4Al(OH)3

..........................3/4 mol...x mol

Na + H2O ----> NaOH + 1/2H2

x mol.......... x mol. 1/2x mol

NaOH + Al(OH)3 -> NaAlO2 + 2H2O

x mol.. x mol

=> 3/4x + 1/2x = 3.36/22.4=0,15 mol

-> x= 0,12 mol -> m= 0,12 * 23 = 2,76g => Đap an B