\(n_{KOH}=0,09.06=0,054\left(mol\right)\\KOH+HCl\rightarrow KCl+H_2O\\ n_{HCl}=n_{KOH}=0,054\left(mol\right)\)

Ủa cho 0,1M chi ta...Hay là đề hỏi thể tích dd HCl em ơi?

\(\text{1)}m_{KOH}=40.35\%=14\left(g\right)\\ \rightarrow n_{KOH}=\dfrac{14}{56}=0,25\left(mol\right)\\ PTHH:KOH+HCl\rightarrow KCl+H_2O\\ \text{Theo pthh}:n_{HCl}=n_{KOH}=0,25\left(mol\right)\\ \rightarrow V_{ddHCl}=0,25.0,5=0,125\left(l\right)\)

\(\text{2)}n_{Al}=\dfrac{4,05}{27}=0,15\left(mol\right)\\ n_{H_2SO_4}=200.14,7\%=29,4\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\\ \text{PTHH}:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\\ \text{LTL}:\dfrac{0,15}{2}< \dfrac{0,3}{3}\rightarrow H_2SO_4\text{ dư}\)

\(\text{Theo pthh}:\left\{{}\begin{matrix}n_{H_2SO_4\left(pư\right)}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,15=0,225\left(mol\right)\\n_{H_2}=n_{H_2SO_4\left(pư\right)}=0,225\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,15=0,075\left(mol\right)\end{matrix}\right.\\ \rightarrow m_{dd\left(\text{sau phản ứng}\right)}=200+4,05-0,3.2=203,45\left(g\right)\)

\(\rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4\text{ dư}}=\dfrac{\left(0,3-0,225\right).98}{203,45}=3,61\%\\C\%_{Al_2\left(SO_4\right)_3}=\dfrac{342.0,075}{203,45}=12,61\%\end{matrix}\right.\)

a) \(n_{KOH}=0,1.1=0,1\left(mol\right)\)

PTHH: 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

Mol:      0,1          0,1            0,1

b) \(V_{ddH_2SO_4}=\dfrac{0,1}{0,5}=0,2\left(l\right)\)

c) \(C_{M_{ddK_2SO_4}}=\dfrac{0,1}{0,1+0,2}=0,333M\)

a) 2NaOH + H₂SO₄→ Na₂SO₄ + 2H₂O 

b) n_NaOH = C_MNaOH . V= 1. 0,1= 0,1mol

PTHH:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

     2             1                 1             2

   0,1         0.05            0,05         0,1

V_H2SO4 =  =0,1l

c) V_{dd sau phản ứng}  = 0,1+0,1=0,2l

C_M =  = 0,25M

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

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

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

\(a/KOH+HCl\rightarrow KCl+H_2O\\ n_{HCl}=0,25.1,5=0,375mol\\ n_{KOH}=n_{KCl}=n_{HCl}=0,375mol\\ V_{KOH}=\dfrac{0,375}{2}=0,1875l\\ b/C_{M_{KCl}}=\dfrac{0,375}{0,1875+0,25}=\dfrac{6}{7}M\)

Chọn B

a) trong 100 ml dung dịch HCl và H2SO4

CM[H+]=[Cl-]=0,02 M

 [SO4 2-]=0,01M

[H+] =2.0,01=0,02 M

trong 100ml dung dịch KOH và Ba(OH)2

[K+]=[OH-]=0,01M

[Ba2+]=0,01M

[OH-]=0,02M

b)n(H+)=0,02+0,02=0,04mol

n(OH-)=0,01+0,02=0,03mol

khi trộn : H+ +  OH-  =>H2O

          0,03<--0,03

=> nH+ dư=0,01mol

=> [H+]=0,05M

=> pH=-lg(0,05)=1,3