`#DyHungg`
Ta có pt:
`2Na + 2H_{2}O -> 2NaOH + H_{2} (1)`
`Na_{2}O + H_{2}O -> 2NaOH (2)`
` n_{H2}=(1,12)/(22,4)=0,05(mol)`
`⇒n_{Na}=0,05xx2=0,1(mol)`
`⇒m_{Na}=0,1xx23=2,3(g)`
`⇒m_{Na2O]=6,95-2,3=4,65 (g)`
`⇒n_{Na2O}=(4,65)/62=0,075 (mol)`
`⇒n_{NaOH}=n_{NaOH (1)} + n_{NaOH (2)}`
`⇒n_{NaOH}=0,1+(0,075xx2)=0,25(mol)`
`⇒C_{M NaOH}=(0,25)/0,5=0,5 (M)`