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

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

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,4}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Fe}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)=3360\left(ml\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{Fe}=0,15\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Fe}=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,3=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{FeCl_2}}=\dfrac{0,15}{0,2}=0,75\left(M\right)\\C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\end{matrix}\right.\)

c, Ta có: \(m_{ddHCl}=1,25.200=250\left(g\right)\)

⇒ m _{dd sau pư} = 8,4 + 250 - 0,15.2 = 258,1 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,15.127}{258,1}.100\%\approx7,38\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{258.1}.100\%\approx1,41\%\end{matrix}\right.\)

Câu 2:

\(m_{H_2SO_4}=39,2.25\%=9,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

\(m_{NaOH}=16.30\%=4,8\left(g\right)\Rightarrow n_{NaOH}=\dfrac{4,8}{40}=0,12\left(mol\right)\)

PTHH: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Mol:      0,06         0,12           0,06

Ta có: \(\dfrac{0,1}{1}>\dfrac{0,12}{2}\) ⇒ H₂SO₄ dư, NaOH hết

m_{dd sau pứ} = 39,2+16 = 55,2 (g)

\(C\%_{ddH_2SO_4dư}=\dfrac{\left(0,1-0,06\right).98.100\%}{55,2}=7,1\%\)

\(C\%_{ddNa_2SO_4}=\dfrac{0,06.142.100\%}{55,2}=15,43\%\)

Câu 3.

a)\(n_{Ba}=\dfrac{13,7}{137}=0,1\left(mol\right)\)

PTHH: Ba + 2H₂O → Ba(OH)₂ + H₂

Mol:     0,1                    0,1           0,1

m_{dd sau pứ} = 13,7+250.1,008-0,1.2 = 265,5 (g)

\(C\%_{ddBa\left(OH\right)_2}=\dfrac{0,1.171.100\%}{265,5}=6,44\%\)

Câu 1 : 

Khối lượng dung dịch là : \(m_{ct}+170\)

Gọi khối lượng muối \(NaNO_3\)cần dùng là x

Ta có :\(m_{ct}=\frac{C\%.m_{dd}}{100}\)

hay \(x=\frac{15.\left(x+170\right)}{100}\)

Ta tính được x=30 (g)

Vậy khối lượng \(NaNO_3\)cần lấy là 30 g

Câu 2 :

 Số mol \(CuSO_4.5H_2O\)là :

\(n_{CuSO_4.5H_2O}=\frac{m}{M}=\frac{75}{250}=0,3\left(mol\right)\)

\(\Rightarrow n_{CuSO_4}=0,3\left(mol\right);V_{dd}=900ml=0,9l\)

\(C_{M_{dd}}=\frac{n}{V}=\frac{0,3}{0,9}=\frac{1}{3}M\)

Vậy...

a, \(C\%_{KCl}=\dfrac{20}{20+60}.100\%=25\%\)

b, \(C\%=\dfrac{40}{40+150}.100\%\approx21,05\%\)

c, \(C\%_{NaOH}=\dfrac{60}{60+240}.100\%=20\%\)

d, \(C\%_{NaNO_3}=\dfrac{30}{30+90}.100\%=25\%\)

e, \(m_{NaCl}=150.60\%=90\left(g\right)\)

f, \(m_{ddA}=\dfrac{25}{10\%}=250\left(g\right)\)

g, \(n_{NaOH}=120.20\%=24\left(g\right)\)

Gọi: n_{NaOH (thêm vào)} = a (g)

\(\Rightarrow\dfrac{a+24}{a+120}.100\%=25\%\Rightarrow a=8\left(g\right)\)

\(a.C_M=\dfrac{0,06}{1,5}=0,04M\\ b.C_M=\dfrac{\dfrac{400}{160}}{4}=0,625M\\ c.C_M=\dfrac{\dfrac{10,53}{58,5}}{\dfrac{450}{1,25}:1000}=0,5M\\ d.C_M=\dfrac{\dfrac{70,2}{40}}{0,5}=3,51M\\ e.C_M=\dfrac{\dfrac{42}{200}}{\dfrac{742}{1,3}:1000}=0,368M\)