BT1:

\(m_{NaCl}=50.20\%+150.10\%=25\left(g\right)\)

\(m_{ddNaCl}=50+150=200\left(g\right)\)

\(C\%_{ddNaCl}=\dfrac{25.100\%}{200}=12,5\%\)

BT2:

\(n_{H_2SO_4}=0,2.5+0,2.3=1,6\left(mol\right)\)

\(V_{ddH_2SO_4}=0,2+0,2=0,4\left(l\right)\)

\(C_{M_{ddH_2SO_4}}=\dfrac{1,6}{0,4}=4M\)

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)

a, \(n_{KOH}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\)

b, \(n_{KOH\left(trong300mlA\right)}=0,3.0,3=0,09\left(mol\right)\)

Gọi: V_H2O = a (l)

\(\Rightarrow C_{M_A}=0,2=\dfrac{0,09}{a+0,3}\Rightarrow a=0,15\left(l\right)=150\left(ml\right)\)

\(n_{Mg}=\dfrac{7.2}{24}=0.3\left(mol\right)\)

\(n_{HCl}=\dfrac{200\cdot14.6\%}{36.5}=0.8\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(0.3.........0.6.........0.3..........0.3\)

\(n_{HCl\left(dư\right)}=0.8-0.6=0.2\left(mol\right)\)

\(KOH+HCl\rightarrow KCl+H_2O\)

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

\(V_{dd_{KOH}}=\dfrac{0.2}{2}=0.1\left(l\right)\)

\(m_{\text{dung dịch sau phản ứng}}=7.2+200-0.3\cdot2=206.6\left(g\right)\)

\(m_{MgCl_2}=0.3\cdot95=28.5\left(g\right)\)

\(C\%MgCl_2=\dfrac{28.5}{206.6}\cdot100\%=13.8\%\)

\(C\%HCl\left(dư\right)=\dfrac{0.2\cdot36.5}{206.6}\cdot100\%=3.53\%\)

Câu 1:

mNaCl= 30*20 /100= 6

a, m dung dịch sau phản ứng là: 30+20= 50

=> C%NaCl= 6/50 *100= 12%

b, m dung dịch còn= 25

=> C% NaCl= 6/25 *100= 24%

Câu 1:

K/l NaCl:

mNaCl= 30 . 20%=6(g)

a,Theo ĐLBTKL:

......30+20=50(g)

=> C%NaCl= \(\frac{6}{50}.100=12\left(\%\right)\)

b, m dung dịch còn= 50-25=25(g)

=> C% NaCl= \(\frac{6}{25}.100=24\left(\%\right)\)

#Walker

Câu 1:

PT ion: \(H^++OH^-\rightarrow H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{OH^-}=0,6\cdot0,4+0,6\cdot0,3\cdot2=0,6\left(mol\right)\\n_{H^+}=0,2\cdot2,6=0,52\left(mol\right)\end{matrix}\right.\) 

\(\Rightarrow\) H⁺ hết, OH^- còn dư \(\Rightarrow n_{OH^-\left(dư\right)}=0,08\left(mol\right)\)

\(\Rightarrow\left[OH^-\right]=\dfrac{0,08}{0,6+0,2}=0,1\left(M\right)\) \(\Rightarrow pH=14+log\left(0,1\right)=13\)

Bài 2:

PT ion: \(H^++OH^-\rightarrow H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{OH^-}=0,3\cdot1,6=0,48\left(mol\right)\\n_{H^+}=0,2\cdot1\cdot2+0,2\cdot2=0,8\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\) OH^- hết, H⁺ còn dư \(\Rightarrow n_{H^+\left(dư\right)}=0,32\left(mol\right)\)

\(\Rightarrow\left[H^+\right]=\dfrac{0,32}{0,2+0,3}=0,64\left(M\right)\) \(\Rightarrow pH=-log\left(0,64\right)\approx0,19\)

a) \(n_{Na}=\dfrac{11,5}{23}=0,5\left(mol\right)\)

\(n_{NaOH}=\dfrac{8\%.500}{40}=1\left(mol\right)\)

\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

0,5---------------->0,5------->0,25

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

\(m_{ddsaupu}=11,5+500-0,25.2=511\left(g\right)\)

=> \(C\%_{NaOH}=\dfrac{1,5.40}{511}.100=11,74\%\)

b) Gọi thể tích dung dịch X cần tìm là V

 \(n_{H^+}=V.1+V.0,5.1=2V\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

Ta có : \(n_{H^+}=n_{OH^-}=1,5\left(mol\right)\)

=> 2V=1,5

=> V=0,75(lít)