em cần gấp lắm ak

W

whitetigerbaekho

NaOH+H2SO4->NaHSO4+H2O
0,3------0,3--------0,3
2NaOH+H2SO4->Na2SO4+2H2O
0,04-----0,02--------0,02
Tổng nNaOH=0,34
mNaOH=13,6
m1=13,6*100/10=136gam
Tổng nH2SO4=0,32
mH2SO4=31,36 gam
m2=31,36*100/18=174,22 gam
 
S

soicon_boy_9x

$H_2SO_4+NaOH \rightarrow NaHSO4+H2O$

$H_2SO_4+2NaOH \rightarrow Na_2SO_4+2H_2O$

Theo bài ra

$n_{NaHSO4}=\dfrac{36}{120}=0,3(mol)$

$n_{Na_2SO_4}=\dfrac{2,84}{142}=0,02(mol)$

Theo pthh:

$n_{H_2SO_4}=0,3+0,02=0,32(mol$

$\rightarrow m_{H_2SO_4}=31,36(g) \rightarrow
m_2=31,36.\dfrac{100}{18}=\dfrac{1568}{9}(g)$

$n_{NaOH}=0,31(mol)$

$\rightarrow m_{NaOH}=12,4(g) \rightarrow m_1=12,4.10=124(g)$




 
W

whitetigerbaekho

soicon_boy_9x said:
$H_2SO_4+NaOH \rightarrow NaHSO4+H2O$

$H_2SO_4+2NaOH \rightarrow Na_2SO_4+2H_2O$

Theo bài ra

$n_{NaHSO4}=\dfrac{36}{120}=0,3(mol)$

$n_{Na_2SO_4}=\dfrac{2,84}{142}=0,02(mol)$

Theo pthh:

$n_{H_2SO_4}=0,3+0,02=0,32(mol$

$\rightarrow m_{H_2SO_4}=31,36(g) \rightarrow
m_2=31,36.\dfrac{100}{18}=\dfrac{1568}{9}(g)$

$n_{NaOH}=0,31(mol)$

$\rightarrow m_{NaOH}=12,4(g) \rightarrow m_1=12,4.10=124(g)$
Bấm để xem đầy đủ nội dung ...

Bạn ơi bạn xem lại bài đi
Bài bạn tính sai rồi
------------------------------------
 
You must log in or register to reply here.
Top Bottom