^mchất tan H₂SO₄ = \(\dfrac{2\cdot19,6\%}{100\%}=0,392\left(g\right)\)

ⁿH₂SO₄ = \(\dfrac{0,392}{2\cdot1+32+4\cdot16}=0,004\left(mol\right)\)

^mchất tan BaCl₂ = \(\dfrac{200\cdot5,2\%}{100\%}=10,4\left(g\right)\)

ⁿBaCl₂ = \(\dfrac{10,4}{137+2\cdot35,5}=0,05\left(mol\right)\)

H₂SO₄ + BaCl₂ \(\rightarrow\) BaSO₄ \(\downarrow\) + 2HCl

1 : 1 : 1 : 2

0,004 \(\rightarrow\) 0,004 \(\rightarrow\) 0,004 \(\rightarrow\) 0,008

Ta có : ⁿH₂SO₄ : \(\dfrac{0,004}{1}\) < ⁿBaCl₂ : \(\dfrac{0,05}{1}\)

\(\Rightarrow\) Tính theo H₂SO₄, BaCl₂ dư

a) ^mkết tủa = 0,004 . (137 + 32 + 4 . 16 ) = 0,932 (g)

b) Các chất sau phản ứng là BaCl₂ dư và HCl

^mdung dịch sau phản ứng = 2 + 200 - 0,932 = 201, 068 (g)

^C%BaCl₂ dư = \(\dfrac{\left(0,05-0,004\right)\cdot\left(137+2\cdot35,5\right)}{201,068}\cdot100\%=4,758\%\)

^C%HCl = \(\dfrac{0,008\cdot\left(1+35,5\right)}{201,068}\cdot100\%=0,145\%\)

PTHH: \(BaCl_2+H_2SO_4\rightarrow2HCl+BaSO_4\downarrow\) (1)

a) Ta có: \(\left\{{}\begin{matrix}m_{BaCl_2}=150\cdot5,2\%=7,8\left(g\right)\\m_{H_2SO_4}=250\cdot19,6\%=49\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{BaCl_2}=\frac{7,8}{208}=0,0375\left(mol\right)\\n_{H_2SO_4}=\frac{49}{98}=0,5\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\frac{0,0375}{1}< \frac{0,5}{1}\) \(\Rightarrow\) BaCl₂ phản ứng hết, H₂SO₄ còn dư

\(\Rightarrow n_{BaSO_4}=0,0375mol\) \(\Rightarrow m_{BaSO_4}=0,0375\cdot233=8,7375\left(g\right)\)

b) Dung dịch A gồm: \(HCl\) và \(H_2SO_{4\left(dư\right)}\)

PTHH: \(HCl+NaOH\rightarrow NaCl+H_2O\) (2)

\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\) (3)

Theo PTHH (1): \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=2n_{BaCl_2}=0,075mol=n_{HCl\left(2\right)}\\n_{H_2SO_4\left(dư\right)}=0,4625mol=n_{H_2SO_4\left(3\right)}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH\left(2\right)}=0,075mol\\n_{NaOH\left(3\right)}=0,925mol\end{matrix}\right.\) \(\Rightarrow n_{NaOH}=1mol\)

\(\Rightarrow V_{NaOH}=\frac{1}{1,5}\approx0,67\left(l\right)=670\left(ml\right)\)

minh ko gioi hoa

1

mình làm hóa có đc 10 thôi mà

a, \(H_2SO_4+BaCl_2\rightarrow BaSO_{4\downarrow}+2HCl\)

b, Ta có: \(m_{H_2SO_4}=114.20\%=22,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{22,8}{96}=0,2375\left(mol\right)\)

\(m_{BaCl_2}=400.5,2\%=20,8\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2375}{1}>\dfrac{0,1}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,1\left(mol\right)\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{H_2SO_4\left(pư\right)}=n_{BaCl_2}=0,1\left(mol\right)\\n_{HCl}=2n_{BaCl_2}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,2375-0,1=0,1375\left(mol\right)\)

Ta có: m _{dd sau pư} = 114 + 400 - 23,3 = 490,7 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,1375.98}{490,7}.100\%\approx2,75\%\\C\%_{HCl}=\dfrac{0,2.36,5}{490,7}.100\%\approx1,49\%\end{matrix}\right.\)

BaCl₂ + H₂SO₄ \(\rightarrow\)BaSO₄ + 2HCl

n_BaCl2=\(\dfrac{31,2}{208}=0,15\left(mol\right)\)

n_H2SO4=\(\dfrac{200.9,6\%}{98}=0,2\left(mol\right)\)

VÌ 0,2>0,15 nên H₂SO₄ dư 0,05 mol

Theo PTHH ta cso:

n_BaCl2=n_BaSO4=0,15(mol)

2n_BaCl2=n_HCl=0,3(mol)

m_BaSO4=233.0,15=34,95(g)

m_HCl=36,5.0,3=10,95(g)

C% dd HCl =\(\dfrac{10,95}{31,2+200-34,95}.100\%=5,58\%\)

C% dd H₂SO₄=\(\dfrac{0,05.98}{196,25}.100\%=2,5\%\)

BaCl2+H2SO4->BaSO4 +2HCl

a.nBaCl2=\(\dfrac{31,2}{208}=0,15\left(mol\right)\)

nH2SO4=\(\dfrac{200\cdot9,6}{100\cdot98}\approx0,2\left(mol\right)\)

Xét tỉ lệ:\(\dfrac{nBaCl2}{nBaCl2pt}=\dfrac{0,15}{1}< \dfrac{nH2SO4}{nH2SO4pt}=\dfrac{0,2}{1}\)

=>BaCl2 hết, Sản phẩm tính theo H2SO4.

nBaSO4=0,15(mol)=>mBaSO4=0,15*233=34,95(g)

b.dd sau pư gồm: H2SO4 dư và HCl

nH2SO4 dư=0,2-0,15=0,05(mol)=>mH2SO4 dư=0,05*98=4,9(g)

nHCl=0,15*2=0,3(mol)=>mHCl=0,3*36,5=10,95(g)

mdd sau pư=31,2+200-34,95=196,25(g)

C%H2SO4=\(\dfrac{4,9\cdot100}{196,25}=2,497\%\)

C%HCl=\(\dfrac{10,95\cdot100}{196,25}=5,58\%\)

gấp á mn giúp em với ;-;