\(m_{KOH.5\%}=30\times5\%=1,5\left(g\right)\)

\(\Rightarrow n_{KOH.5\%}=\frac{1,5}{56}=\frac{3}{112}\left(mol\right)\)

\(m_{KOH.15\%}=20\times15\%=3\left(g\right)\)

\(\Rightarrow n_{KOH.15\%}=\frac{3}{56}\left(mol\right)\)

Ta có: \(m_{ddKOH}mới=30+20=50\left(g\right)\)

\(m_{KOH}mới=1,5+3=4,5\left(g\right)\)

\(\Rightarrow C\%_{KOH}mới=\frac{4,5}{50}\times100\%=9\%\)

Ta có: \(V_{ddKOH}mới=\frac{50}{1,1}=45,45\left(ml\right)=0,04545\left(l\right)\)

\(n_{KOH}mới=\frac{3}{112}+\frac{3}{56}=\frac{9}{112}\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}mới=\frac{9}{112}\div0,04545=1,77\left(M\right)\)

Khối lượng KOH trong dung dịch tạo thành sau khi trộn

\(m_{KOH}=30.\frac{5}{100}+20.\frac{15}{100}=45\left(g\right)\approx\left(0,08mol\right)\)

Thể tích dung dịch tạo thành:

\(V_{dd}=\frac{m_{dd}}{D}=\frac{30+20}{1,1,}=45,45ml=0,045l\)

\(\rightarrow C\%=\frac{m_{ct}}{m_{dd}}.100\%=\frac{4,5}{50}.100\%=9\%\)

\(C_M=\frac{n_{ct}}{V_{dd}}=\frac{0,08}{0,045}=1,77M\)

\(\Sigma m_{KOH}=30.5\%+20.15\%=4,5\left(g\right)\)

=> C% = \(\frac{4,5}{30+20}.100\%=9\%\)

=> n_KOH = 0,08 (mol)

V = m/D = 50/1,1 \(\approx45,455\left(ml\right)\) = 1/22 (l)

C_M = n/V = 1,76 (M)

2

b

mNaCl=\(\dfrac{200.15}{100}\)=30(g)

nNaCl=\(\dfrac{30}{58,5}\)=0.51(mol)

VddNaCl=\(\dfrac{200}{1,1}\)=181.8(ml)=0.1818(l)

CMNaCl=\(\dfrac{0,51}{0,1818}\)=2.8(M)

1) Ta có: \(m_{H_2SO_4}=200\cdot15\%+300\cdot25\%=105\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{105}{200+300}\cdot100\%=21\%\)

2) Ta có: \(\left\{{}\begin{matrix}n_{H_2SO_4}=\dfrac{105}{98}=\dfrac{15}{14}\left(mol\right)\\V_{ddH_2SO_4}=\dfrac{500}{1,25}=400\left(ml\right)\end{matrix}\right.\) \(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{\dfrac{15}{14}}{0,4}\approx2,68\left(M\right)\)

2 dd trên là 2 dd nào vậy ?

\(a,n_{H_2SO_4}=0,3.0,75+0,3.0,25=0,3\left(mol\right)\\ V_{ddH_2SO_4}=300+300=600\left(ml\right)=0,6\left(l\right)\\ \rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{0,3}{0,6}=0,5M\\ m_{H_2SO_4}=0,3.98=29,4\left(g\right)\\ m_{ddH_2SO_4}=600.1,02=612\left(g\right)\\ \rightarrow C\%_{H_2SO_4}=\dfrac{29,4}{612}.100\%=4,8\%\)

\(b,\) Đặt kim loại M có hoá trị n (n ∈ N*)

PTHH: \(2M+nH_2SO_4\rightarrow M_2\left(SO_4\right)_n+nH_2\uparrow\)

           \(\dfrac{0,6}{n}\)<---0,3--------------------------->0,3

\(\rightarrow M_M=\dfrac{5,4}{\dfrac{0,6}{n}}=9n\left(g\text{/}mol\right)\)

Vì n là hoá trị của M nên ta xét bảng

Vậy M là Al

\(c,n_{KClO_3}=\dfrac{15,3125}{122,5}=0,125\left(mol\right)\)

PTHH:

\(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)

0,3-->0,15

\(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\uparrow\)

0,1<---------------------0,15

\(\rightarrow H=\dfrac{0,1}{0,125}.100\%=80\%\)

bai 1 ne 

\(n_{KOH}=0.1\cdot1=0.1\left(mol\right)\)

\(n_{H_2SO_4}=0.3\cdot0.5=0.15\left(mol\right)\)

\(2KOH+H_2SO_4\rightarrow K_2SO_4+H_2O\)

\(0.1..........0.05...............0.05\)

Dung dịch D : 0.05 (mol) K₂SO₄ , 0.1 (mol) H₂SO₄

\(\left[K^+\right]=\dfrac{0.05\cdot2}{0.1+0.3}=0.25\left(M\right)\)

\(\left[H^+\right]=\dfrac{0.1\cdot2}{0.1+0.3}=0.5\left(M\right)\)

\(\left[SO_4^{2-}\right]=\dfrac{0.05+0.1}{0.1+0.3}=0.375\left(M\right)\)

\(2NaOH+H_2SO_4\rightarrow K_2SO_4+H_2O\)

\(0.2..................0.1\)

\(V_{dd_{NaOH}}=\dfrac{0.2}{1}=0.2\left(l\right)\)