Đáp án:

`a)\ m_{Na_2CO_3}=36(g)`

`b)\ m=4,6a(g)\ ; V=2,24a(l)`

Giải thích các bước giải:

`a)`

`m_{Na_2CO_3}=200.18%=36(g)`

`b)`

`Pt : 2Na+2H_2O->2NaOH+H_2`

`Gia\ su: C_{M\ dd\ NaOH}=a(M)`

`n_{NaOH}={200}/{1000}.a=0,2a(mol)`

`Theo\ pt: n_{Na}=n_{NaOH}=0,2a(mol)`

`m_{Na}=0,2a.23=4,6a(g)`

`Theo\ pt: n_{H_2}=1/{2}n_{NaOH}=0,1a(mol)`

`V_{H_2}=0,1a.22,4=2,24a(l)`

$\text{Đáp án + giải thích các bước giải :}$

$Bài$ $1$ $:$ 

$a,$

$m_{Na_2CO_3}$ $=$ $m_{\text{dung dịch Na_2CO_3}}$ $×$ $C\%_{Na_2CO_3}$ $=$ $200$ $×$ $18\%$ $=$ $36$ $(g)$

$⇒$ $\text{Cần 36 gam}$ $Na_2CO_3$ 

$b,$

$PTHH$ $:$ 

$2Na$ $+$ $2H_2O$ $→$ $2NaOH$ $+$ $H_2↑$

$n_{Na}$ $=$ `m/23` $(mol)$

$Theo$ $pt$ $:$ $n_{H_2}$ $=$ `1/2`$n_{Na}$ $=$ `m/46` $(mol)$

$⇒$ $V_{H_2}$ $=$ `m/46` $×$ $22,4$ $=$ `(56m)/115` $(l)$

$⇒$ `(56m)/115` $=$ $V$

$\text{Chúc bn hk tốt !! :))))}$