ai giúp mình với, mình đang ktra huhuhuh

\(m_{H_2SO_4}=\dfrac{19,6\cdot20\%}{100\%}=3,92\left(g\right)\\ \Rightarrow n_{H_2SO_4}=\dfrac{3,92}{98}=0,04\left(mol\right)\\ PTHH:H_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2HCl\\ \Rightarrow n_{H_2SO_4}=n_{BaCl_2}=n_{BaSO_4}=0,04\left(mol\right)\\ \Rightarrow m_{CT_{BaCl_2}}=0,04\cdot208=8,32\left(g\right)\\ \Rightarrow m_{dd_{BaCl_2}}=\dfrac{8,32\cdot100\%}{12\%}\approx69,3\left(g\right)\\ m_{kết.tủa}=m_{BaSO_4}=0,04\cdot233=9,32\left(g\right)\)

a) \(\left\{{}\begin{matrix}n_{CuSO_4}=0,3.1=0,3\left(mol\right)\\n_{BaCl_2}=0,1.2=0,2\left(mol\right)\end{matrix}\right.\)

PTHH:           \(CuSO_4+BaCl_2\rightarrow BaSO_4\downarrow+CuCl_2\)

Ban đầu:         0,3           0,2

Sau pư:           0,1            0              0,2             0,2

=> \(m_{kt}=m_{BaSO_4}=0,2.233=46,6\left(g\right)\)

b)             \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)

                 0,1-------->0,2

                \(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)

                 0,2------>0,4

=> \(m_{ddNaOH}=\dfrac{\left(0,2+0,4\right).40}{15\%}=160\left(g\right)\)

\(n_{BaCl_2}=\dfrac{208.10\%}{208}=0,1\left(mol\right)\\ a,BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ 0,1............0,1..............0,1.............0,2\left(mol\right)\\ b,m_{ddH_2SO_4}=\dfrac{0,1.98.100}{8}=122,5\left(g\right)\\ c,m_{kt}=m_{BaSO_4}=0,1.233=23,3\left(g\right)\\ d,m_{ddsau}=208+122,5-23,3=307,2\left(g\right)\\ C\%_{ddHCl}=\dfrac{0,2.36,5}{307,2}.100\approx2,376\%\)

a)

$BaCl_2 + H_2SO_4 \to BaSO_4 + 2HCl$
$n_{BaCl_2} = 0,1 < n_{H_2SO_4} = 0,2$ nên $H_2SO_4$ dư

$n_{BaSO_4} = n_{BaCl_2} = 0,1(mol)$
$m_{BaSO_4} = 0,1.233 = 23,3(gam)$

b)

A gồm :

$HCl : 0,1.2 = 0,2(mol)$
$H_2SO_4\ dư : 0,2 - 0,1 = 0,1(mol)$
$V_{dd} = 0,1 + 0,1= 0,2(lít)$

$C_{M_{HCl}} = \dfrac{0,2}{0,2} = 1M$
$C_{M_{H_2SO_4}} = \dfrac{0,1}{0,2} = 0,5M$

c)

$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} = 2n_{H_2SO_4\ dư} = 0,2(mol)$
$m_{dd\ NaOH} = \dfrac{0,2.40}{15\%} = 53,33(gam)$

m_ddH2SO4 = 100 . 1,137 = 113,7

n_H2SO4 = 113,7 . 20%/98 = 0,232 mol

n_BaCl2 = 400 . 5,29%/208 = 0,1 mol

                 H₂SO₄ + BaCl₂ —> BaSO₄ + 2HCI

Bđ:            0,232      0,1

Pứ:            0,1        0, 1         0,1       0,2

Sau pứ:      0,132         0

m_BaSO4 = 0,1.233 = 23,3 gam

Khối lượng dung dịch sau khi lọc bỏ kết tủa:

m_ddB = m_ddH2SO4 + m_ddBaCl2 - m_BaSO4 = 490,4

C%HCI = 0,2.36,5/490,4 = 1,49%

C%_H2SO4 dư = 0,132.98/490,4 = 2,64%

`a)`

`n_{SO_3}=0,005(mol)`

`m_{H_2SO_4\ 10\%}=10\%.a=0,1a(g)`

`SO_3+H_2O->H_2SO_4`

`0,005->0,005->0,005(mol)`

`->C\%_{H_2SO_4\ sau}={0,005.98+0,1a}/{0,4+a}.100=12,25`

`->a=19,6(g)`

`b)`

`n_{Ba(OH)_2}=0,01.0,5=0,005(mol)`

`n_{H_2SO_4}=0,005+{0,1.19,6}/{98}=0,025(mol)`

`n_{NaOH}=0,05.0,8=0,04(mol)`

`Ba(OH)_2+H_2SO_4->BaSO_4+2H_2O`

`0,005->0,005(mol)`

`->n_{H_2SO_4\ du}=0,025-0,005=0,02(mol)`

`2NaOH+H_2SO_4->Na_2SO_4+2H_2O`

Do `0,02={0,04}/2->` Pu hoàn toàn.

Theo PT : `n_{Na_2SO_4}=n_{H_2SO_4}=0,02(mol)`

`->m_{Na_2SO_4}=0,02.142=2,84<6,44`

`->X` là muối ngậm nước.

Đặt X là `Na_2SO_4.xH_2O`

Bảo toàn Na: `n_{Na_2SO_4.xH_2O}=n_{Na_2SO_4}=0,02(mol)`

`->M_{Na_2SO_4.xH_2O}={6,44}/{0,02}=322(g//mol)`

`->142+18x=322`

`->x=10`

`->X` là `Na_2SO_4.10H_2O`

nH2SO4=0,02.1=0,02(ol)

a) PTHH: 2 NaOH + H2SO4 -> Na2SO4 + 2 H2O

0,04____________0,02____0,02(mol)

mNaOH=0,04.40= 1,6(g)

=>mddNaOH= (1,6.100)/20= 8(g)

b) PTHH: H2SO4 + 2 KOH -> K2SO4 + 2 H2O

0,2____________0,04(mol)

=>mKOH=0,04.56=2,24(g)

=>mddKOH= (2,24.100)/5,6=40(g)

=>VddKOH= mddKOH/DddKOH= 40/1,045=38,278(ml)