Bài 1:

PTHH: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Ta có: \(\left\{{}\begin{matrix}m_{Ba\left(OH\right)_2}=150\cdot17,1\%=25,65\left(g\right)\\m_{HCl}=300\cdot7,3\%=21,9\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{Ba\left(OH\right)_2}=\frac{25,65}{171}=0,15\left(mol\right)\\n_{HCl}=\frac{21,9}{36,5}=0,6\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\frac{0,15}{1}< \frac{0,6}{2}\) \(\Rightarrow\) Ba(OH)₂ phản ứng hết, HCl còn dư

\(\Rightarrow\) Dung dịch A làm quỳ tím hóa đỏ

Bài 2:

PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=\frac{150\cdot4\%}{40}=0,15\left(mol\right)\\n_{H_2SO_4}=\frac{300\cdot9,8\%}{98}=0,3\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\frac{0,15}{2}< \frac{0,3}{1}\) \(\Rightarrow\) NaOH phản ứng hết, H₂SO₄ còn dư

\(\Rightarrow\) Dung dịch A làm quỳ tím hóa đỏ

Bài 3:

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

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

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 chứa \(HCl\) 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): \(\left\{{}\begin{matrix}n_{HCl}=2n_{BaCl_2}=0,075mol\\n_{H_2SO_4\left(dư\right)}=0,4625mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{NaOH\left(2\right)}=0,075mol\\n_{NaOH\left(3\right)}=0,925mol\end{matrix}\right.\)

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

minh ko gioi hoa

1

mình làm hóa có đc 10 thôi mà

a) \(2NaOH+H2SO4--->Na2SO4+2H2O\) (1)

\(Ba\left(OH\right)2+H2SO4--->BaSO4+2H2O\)

nBaSO4 = 18,64/233 = 0,08 mol

nH2SO4 cần dùng = 0,07 . 2 = 0,14 mol

- Theo PTHH (2): nH2SO4 = 0,08 mol

=> nH2SO4 (1) = 0,14 - 0,08 = 0,06 mol

=> nBa(OH)2 = nH2SO4 (2) = 0,08 mol

=> CM Ba(OH)2 = 0,08/ 0,2 = 0,4M

=> nNaOH = nH2SO4 (1) = 0,12 mol

=> CM NaOH = 0,12/0.2 = 0,6M

Bài 19  :

\(a) n_{Al} = \dfrac{10,8}{27} = 0,4(mol)\\ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{H_2} = \dfrac{3}{2}n_{Al} = 0,6(mol)\\ V_{H_2} = 0,6.22,4 = 13,44(lít)\\ b) \text{Chất tan : }Al_2(SO_4)_3\\ n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,2(mol)\\ m_{Al_2(SO_4)_3} = 0,2.342 = 68,4(gam)\)

Bài 18 : 

\(a) n_{HCl} = \dfrac{250.7,3\%}{36,5 } = 0,5(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,25(mol) \Rightarrow V_{H_2} = 0,25.22,4 = 5,6(lít)\\ b) \text{Chất tan : } ZnCl_2\\ n_{ZnCl_2} = n_{H_2} = 0,25(mol)\\ m_{ZnCl_2} = 0,25.136 = 34(gam)\)

Bài 1:

a) CuSO₄ + 2NaOH → Na₂SO₄ + Cu(OH)₂↓

\(n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)

\(n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\)

Theo PT: \(n_{CuSO_4}=\dfrac{1}{2}n_{NaOH}\)

Theo bài: \(n_{CuSO_4}=\dfrac{1}{3}n_{NaOH}\)

Vì \(\dfrac{1}{3}< \dfrac{1}{2}\) ⇒ NaOH dư

b) Theo PT: \(n_{Cu\left(OH\right)_2}=m_{CuSO_4}=0,1\left(mol\right)\)

\(\Rightarrow m_{Cu\left(OH\right)_2}=0,1\times98=9,8\left(g\right)\)

c) \(\Sigma V_{dd}saupư=40+60=100\left(ml\right)=0,1\left(l\right)\)

Theo PT: \(n_{NaOH}pư=2n_{CuSO_4}=2\times0,1=0,2\left(mol\right)\)

\(\Rightarrow n_{NaOH}dư=0,3-0,2=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}dư=\dfrac{0,1}{0,1}=1\left(M\right)\)

Theo PT: \(n_{Na_2SO_4}=n_{CuSO_4}=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,1}{0,1}=1\left(M\right)\)

Bài 2:

ZnCl₂ + 2NaOH → 2NaCl + Zn(OH)₂↓ (1)

\(n_{ZnCl_2}=0,3\times1,5=0,45\left(mol\right)\)

\(n_{NaOH}=0,1\times1=0,1\left(mol\right)\)

Theo PT1: \(n_{ZnCl_2}=\dfrac{1}{2}n_{NaOH}\)

Theo bài: \(n_{ZnCl_2}=\dfrac{9}{2}n_{NaOH}\)

Vì \(\dfrac{9}{2}>\dfrac{1}{2}\) ⇒ ZnCl₂ dư

a) \(\Sigma V_{dd}saupư=300+100=400\left(ml\right)=0,4\left(l\right)\)

Theo PT1: \(n_{ZnCl_2}pư=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\times0,1=0,05\left(mol\right)\)

\(\Rightarrow n_{ZnCl_2}dư=0,45-0,05=0,4\left(mol\right)\)

\(\Rightarrow C_{M_{ZnCl_2}}dư=\dfrac{0,4}{0,4}=1\left(M\right)\)

Theo PT1: \(n_{NaCl}=n_{NaOH}=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\)

b) Zn(OH)₂ \(\underrightarrow{to}\) ZnO + H₂O (2)

Theo pT1: \(n_{Zn\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\times0,1=0,05\left(mol\right)\)

Theo pT2: \(n_{ZnO}=n_{Zn\left(OH\right)_2}=0,05\left(mol\right)\)

\(\Rightarrow m_{ZnO}=0,05\times81=4,05\left(g\right)\)

c) NaOH + HCl → NaCl + H₂O (3)

Theo PT: \(n_{HCl}=n_{NaOH}=0,1\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,1\times36,5=3,65\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{3,65}{25\%}=14,6\left(g\right)\)

1.

\(nOH^-=2nBa\left(OH\right)_2+nKOH=2.0,25.0,01+0,25.0,02=0,01mol\)\(nH^+=2nH_2SO_4=0,5a\left(mol\right)\)

Dung dịch sau phản ứng là môi trường axit.

\(pH=2\Rightarrow\left[H^+\right]=10^{-2}M\)

\(\frac{nH^+-nOH^-}{V}=\left[H^+\right]\)

\(\Leftrightarrow\frac{0,5a-0,01}{0,5}=10^{-2}\)

\(\Leftrightarrow a=0,03M\)

\(nBa^{2+}=2,5.10^{-3}mol\)

\(nSO_4^{2-}=7,5.10^{-3}mol\)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)

Chất sản phẩm tính theo nBa²⁺

\(b=2,5.10^{-3}.233=0,5825g\)

Không biết có đúng hay không.

2.Gọi A là thể tích dd NaOH (lít)

\(nAl^{3+}=2nAl_2\left(SO_4\right)_3=2.0,1.0,05=0,01mol\)

\(nOH^-=nNaOH=0,2A\left(mol\right)\)

\(nAl\left(OH\right)_3=\frac{0,78}{78}=0,01mol\)

\(Al^{3+}+3OH^-\rightarrow Al\left(OH\right)_3\)

\(0,01-0,6A---0,01--mol\)

Ta có:

\(\frac{nAl^{3+}}{1}=\frac{nOH^-}{3}\)

\(\Leftrightarrow0,01=\frac{0,6A}{3}\)

\(\Leftrightarrow A=0,05=50ml\)

đang cần gấp ai lm đc gúp em trong chiều nay đc ko ạ cảm ơn nhìu nhìu

CaCl₂ trộn với NaOH không tạo kết tủa nha em!