\(\left\{{}\begin{matrix}n_{HCl}=0,1.1=0,1\left(mol\right)\\n_{H_2SO_4}=0,1.1=0,1\left(mol\right)\end{matrix}\right.\)

PTHH: 

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

0,1<------0,1

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

0,2<--------0,1

\(\Rightarrow V_{ddNaOH}=\dfrac{0,2+0,1}{1}=0,3\left(l\right)\)

\(n_{HCl}=0,1.1=0,1\left(mol\right);n_{H_2SO_4}=0,1.1=0,1\left(mol\right)\)

PTHH: 

`NaOH + HCl -> NaCl + H_2O`

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

Theo PT: `n_{NaOH} = 2n_{H_2SO_4} + n_{HCl} = 0,3 (mol)`

`=> V_{ddNaOH} = (0,3)/(1) = 0,3(l)`

\(n_{HCl}=0,2.2=0,4\left(mol\right)\)

\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\)

PT: \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

\(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)

Theo PT: \(n_{Ca\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}+n_{H_2SO_4}=0,4\left(mol\right)\)

\(\Rightarrow V_{ddCa\left(OH\right)_2}=\dfrac{0,4}{1}=0,4\left(l\right)\)

Đề câu 2 sao khỏi làm đi

\(n_{HCl}=0.1\cdot0.03=0.003\left(mol\right)\)

\(n_{NaOH}=0.1\cdot0.01=0.001\left(mol\right)\)

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

Lập tỉ lệ : 

\(\dfrac{0.003}{1}>\dfrac{0.001}{1}\Rightarrow HCldư\)

\(n_{HCl\left(dư\right)}=0.003-0.001=0.002\left(mol\right)\)

\(\left[H^+\right]=\dfrac{0.002}{0.1+0.1}=0.01\)

\(pH=-log\left(0.01\right)=2\)

\(b.\)

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

\(0.001..........0.002\)

\(V_{Ba\left(OH\right)_2}=\dfrac{0.001}{1}=0.001\left(l\right)\)

\(n_{KOH}=0,2.1=0,2\left(mol\right)\\ 2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\\ KOH+HCl\rightarrow KCl+H_2O\\ n_{KOH}=2.n_{H_2SO_4}+n_{HCl}=2.1.V+0,5.V\\ \Leftrightarrow0,2=2,5.V\\ \Leftrightarrow V=0,08\left(lít\right)\)

Ta có: \(\left\{{}\begin{matrix}n_{HCl}=V\left(mol\right)\\n_{H_2SO_4}=0,5V\left(mol\right)\\n_{KOH}=0,2\left(mol\right)\end{matrix}\right.\)

PT ion: \(H^++OH^-\rightarrow H_2O\)

\(\Rightarrow\) Dung dịch sau p/ứ gồm các ion Cl^- (V mol), SO₄^2- (0,5V mol), K⁺ (0,2 mol)

Bảo toàn điện tích: \(n_{Cl^-}+2n_{SO_4^{2-}}=n_K\) \(\Rightarrow V+2\cdot0,5V=0,2\)

\(\Rightarrow V=0,1\left(l\right)=100\left(ml\right)\)

Anh tính toán có nhầm lẫn, em tham khảo cách của bạn Duy Anh nha!

1 ) nOH= nNaOH+2nBa(OH)2=0,01+2.0,01=0,03mol 
OH- + H+ ----------->H2O 
0.03-->0.03 
nHCl=nH+ =0.03mol 
=>VHCl=0.03/0.3=0.1l=100ml

2 )  (NH4)2SO4 + 2KOH -(t°)-> K2SO4 + 2NH3 + 2H2O 
0,05______________________________ 0,1 (mol) 
n(NH4)2SO4 = 0,05.1 = 0,05 mol 
V NH3 = 0,1 . 22,4 = 2,24 l 

\(n_{HCl}=0,5.1=0,5\left(mol\right)\)

\(n_{H2SO4}=0,5.1=0,5\left(mol\right)\)

Ta có PTHH:

\(2NaOH+HCl\underrightarrow{ }NaCl+2H_{2_{ }}O\)

\(n_{NaOH}=n_{HCl}+n_{H2SO4}=0,5+0,5=1\left(mol\right)\)

\(\Rightarrow V_{NaOH}=1:1=1\left(lít\right)\)

Nhìn theo bài gần giống rồi làm, chúc bạn học tốt !

\(n_{HCl}=0,1.1=0.1\left(mol\right)\)

\(n_{H2SO\text{4}}=0,1.1=0,1\left(mol\right)\)

Pthh \(NaOH+HCl-->NaCl+H_2O\)

\(2NaOH+H2SO4-->Na2SO4+2H2O\)\(=>n_{NaOH}=0,1+0,1.2=0,3\left(mol\right)\)

\(=>V_{NaOH}=\dfrac{0,3}{1}=0,3\left(l\right)\)

\(NaOH+HCl\xrightarrow[]{}NaCl+H_2O\)( phương trình 1)

\(2NaOH+H_{2_{ }}SO_4\xrightarrow[]{}Na_{2_{ }}SO_4+2H_{2_{ }}O\)( phương trình 2)

Theo bài ta có:

\(n_{HCl}=C_M.V=1.0,1=0,1\left(mol\right)\)

\(n_{H2SO4}=C_M.V=1.0,1=0,1\left(mol\right)\)\(\Rightarrow n_{NaOH\left(1\right)}=0,1\left(mol\right)\)

\(n_{NaOH\left(2\right)}=0,1.2=0,2\left(mol\right)\Rightarrow n_{NaOH}=0,1+0,2=0,3\left(mol\right)\)

\(\Rightarrow V_{NaOH}=\dfrac{n_{NaOH}}{C_M}=\dfrac{0,3}{1}=0,3\left(lít\right)\)

Cô ơi, bài em làm lại, cô xem hộ xem có đúng không nha! Cẩm Vân Nguyễn Thị

\(n_{NaOH\left(dư\right)}=n_{HCl}=0.1\cdot1=0.1\left(mol\right)\)

\(\Rightarrow n_{NaOH\left(pư\right)}=0.6-0.1=0.5\left(mol\right)\)

\(n_{H_2O}=n_{NaOH}=0.5\left(mol\right)\)

\(BTKL:\)

\(m_{Muối}=24.2+0.5\cdot40-0.5\cdot18=35.2\left(g\right)\)