\(n_{H_2SO_4}=0,4\cdot1=0,4mol\)

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

 0,8              0,4

\(V_{NaOH}=\dfrac{0,8}{0,5}=1,6l\)

\(n_{H_2SO_4}=0.4\cdot1=0.4\left(mol\right)\)

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

\(0.8.............0.4..............0.4\)

\(V_{dd_{NaOH}}=\dfrac{0.8}{0.5}=1.6\left(l\right)\)

\(V_{dd}=0.4+1.6=2\left(l\right)\)

\(C_{M_{Na_2SO_4}}=\dfrac{0.4}{2}=0.2\left(M\right)\)

\(a,PTHH:KOH+HCl\rightarrow KCl+H_2O\\ b,n_{KOH}=n_{HCl}=2\cdot0,1=0,2\left(mol\right)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,2}{2}=0,1\left(l\right)\\ c,C_{M_{KCl}}=\dfrac{0,2}{0,1+0,1}=1M\)

Đổi 100ml = 0,1 lít

Ta có: \(n_{KOH}=2.0,1=0,2\left(mol\right)\)

a. PTHH: \(KOH+HCl--->KCl+H_2O\)

b. Theo PT: \(n_{HCl}=n_{KOH}=0,2\left(mol\right)\)

\(\Rightarrow V_{dd_{HCl}}=\dfrac{0,2}{2}=0,1\left(lít\right)=100\left(ml\right)\)

c. Ta có: \(V_{dd_{KCl}}=V_{dd_{HCl}}=0,1\left(lít\right)\)

Theo PT: \(n_{KCl}=n_{HCl}=0,2\left(mol\right)\)

\(\Rightarrow C_{M_{KCl}}=\dfrac{0,2}{0,1}=2M\)

bôi đen chi zợ, bôi trắng ik pn

HCl+NaOH->NaCl+H2O

0,2-------0,2----0,2

n HCl=0,2 mol

=>m NaOH=0,2.40=8g

->mdd NaOH=40g

=>m NaCl=0,2.58,5=11,7g

\(NaOH+HCl\rightarrow NaCl+H_2O\)

    0,2        0,2           0,2              ( mol )

\(n_{HCl}=1.0,2=0,2mol\)

\(m_{NaOH}=0,2.40:20\%=40g\)

\(m_{NaCl}=0,2.58,5=11,7g\)

\(a,H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ n_{H_2SO_4}=1.0,4=0,4\left(mol\right)\\ n_{NaOH}=0,4.2=0,8\left(mol\right)\\ b,V_{ddNaOH}=\dfrac{0,8}{0,5}=1,6\left(l\right)\\ c,n_{Na_2SO_4}=n_{H_2SO_4}=0,4\left(mol\right)\\ V_{ddNa_2SO_4}=0,4+1,6=2\left(l\right)\\ C_{MddNa_2SO_4}=\dfrac{0,4}{2}=0,2\left(M\right)\)

Phương trình hoá học :

H 2 SO 4  + 2NaOH →  Na 2 SO 4 + 2 H 2 O

a) Ba(OH)₂ + 2HCl --> BaCl₂ + 2H₂O

b) \(n_{Ba\left(OH\right)_2}=0,5.0,4=0,2\left(mol\right)\)

PTHH: Ba(OH)₂ + 2HCl --> BaCl₂ + 2H₂O

               0,2------>0,4----->0,2

=> \(V_{ddHCl}=\dfrac{0,4}{1,5}=\dfrac{4}{15}\left(l\right)\)

c) \(m_{BaCl_2}=0,2.208=41,6\left(g\right)\)

2HCl+Ba(OH)₂->BaCl₂+2H₂O

0,4------0,2-----------0,2 mol

n _Ba(OH)2=0,5.0,4=0,2 mol

=>V_HCl=\(\dfrac{0,4}{1,5}\)=\(\dfrac{4}{15}l\)

=>m _BaCl2=0,2.208=41,6g

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b, Có lẽ phần này đề hỏi khối lượng sắt chứ bạn nhỉ?

 \(n_{ZnCl_2}=0,4.2=0,8\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{FeCl_2}=0,8\left(mol\right)\Rightarrow m_{Fe}=0,8.56=44,8\left(g\right)\)

c, \(n_{H_2}=n_{FeCl_2}=0,8\left(mol\right)\Rightarrow V_{H_2}=0,8.22,4=17,92\left(l\right)\)

d, \(n_{HCl}=2n_{FeCl_2}=1,6\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{1,6}{0,4}=4\left(M\right)\)

a) PT phân tử: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)

    PT ion: \(CaCO_3+2H^+\rightarrow Ca^{2+}+H_2O+CO_2\uparrow\)

b) Ta có: \(\left\{{}\begin{matrix}n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\\n_{HCl}=\dfrac{43,8\cdot20\%}{36,5}=0,24\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,24}{2}\) \(\Rightarrow\) HCl dư, tính theo CaCO₃

\(\Rightarrow\left\{{}\begin{matrix}n_{CaCl_2}=0,1\left(mol\right)=n_{CO_2}\\n_{HCl\left(dư\right)}=0,04\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCl_2}=0,1\cdot111=11,1\left(g\right)\\m_{CO_2}=0,1\cdot44=4,4\left(g\right)\\m_{HCl\left(dư\right)}=0,04\cdot36,5=1,46\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{CaCO_3}+m_{ddHCl}-m_{CO_2}=49,4\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{CaCl_2}=\dfrac{11,1}{49,4}\cdot100\%\approx22,47\%\\C\%_{HCl\left(dư\right)}=\dfrac{1,46}{49,4}\cdot100\%\approx2,96\%\end{matrix}\right.\)