Cho 4,6g Na vào 100g dd NaOH 5% tính C% của dd NaOH PTHH 2Na + 2H2O —> 2NaOH + H2 25/09/2021 Bởi Eliza Cho 4,6g Na vào 100g dd NaOH 5% tính C% của dd NaOH PTHH 2Na + 2H2O —> 2NaOH + H2
Ban đầu, $m_{NaOH}=100.5\%=5g$ $n_{Na}=\dfrac{4,6}{23}=0,2 mol$ $\Rightarrow n_{NaOH}=0,2 mol; n_{H_2}=0,1 mol$ $m_{NaOH(dd)}=5+0,2.40=13g$ $m_{dd\text{spứ}}=m_{Na}+100-m_{H_2}=4,6+100-0,1.2=104,4g$ $\Rightarrow C\%_{NaOH}=\dfrac{13.100}{104,4}=12,45\%$ Bình luận
PTHH `2Na + 2H_2O -> 2NaOH + H_2` `n_(Na)=\frac{4,6}{23} =0,2(mol)` Theo pthh `n_(NaOH)=n_(Na)=0,2(mol)` `=> m_(NaOH)=0,2.40=8(g)` `n_(H_2)=1/2.n_(Na)=0,1(mol)` `=> m_(H_2)=0,1.2=0,2(g)` `m_(NaOH)(trong dd )= 100.5/100=5(g)` `=> m_(NaOH) sau = 5+8=13(g)` `m_(dd)=4,6+100-0,2=104,4(g)` `=>C%=\frac{13}{104,4} .100=12,45%` Bình luận
Ban đầu, $m_{NaOH}=100.5\%=5g$
$n_{Na}=\dfrac{4,6}{23}=0,2 mol$
$\Rightarrow n_{NaOH}=0,2 mol; n_{H_2}=0,1 mol$
$m_{NaOH(dd)}=5+0,2.40=13g$
$m_{dd\text{spứ}}=m_{Na}+100-m_{H_2}=4,6+100-0,1.2=104,4g$
$\Rightarrow C\%_{NaOH}=\dfrac{13.100}{104,4}=12,45\%$
PTHH
`2Na + 2H_2O -> 2NaOH + H_2`
`n_(Na)=\frac{4,6}{23} =0,2(mol)`
Theo pthh `n_(NaOH)=n_(Na)=0,2(mol)`
`=> m_(NaOH)=0,2.40=8(g)`
`n_(H_2)=1/2.n_(Na)=0,1(mol)`
`=> m_(H_2)=0,1.2=0,2(g)`
`m_(NaOH)(trong dd )= 100.5/100=5(g)`
`=> m_(NaOH) sau = 5+8=13(g)`
`m_(dd)=4,6+100-0,2=104,4(g)`
`=>C%=\frac{13}{104,4} .100=12,45%`