Question

Answer 1

a) ¯¯¯¯¯¯MA=3,30,3=11(g/mol)M¯A=3,30,3=11(g/mol)

0,212>0,0910,212>0,091 => O₂ hết, H₂ dư

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

                    0,09------>0,18

=> mH2O=0,18.18=3,24(g)

Answer 2

Answer 3

nA = 6,72/22,4 = 0,3 (mol)

MA = 3,3/0,3 = 11 (g/mol)

dA/kk = 11/29 = 0,379 (lần)

Gọi nO2 = a (mol); nH2 = b (mol)

Ta có:

a + b = 0,3 (mol)

32a + 2b = 3,3 (g)

<=> a = 0,09 (mol); b = 0,21 (mol)

PTHH: 2H2 + O2 -> (t°) 2H2O

LTL: 0,21/2 > 0,09 => H2 dư

nH2O = 0,09 . 2 = 0,18 (mol)

mH2O = 0,18 . 18 = 3,24 (g)