Bài 1: \(Zn\left(0,04\right)+2HCl\left(0,08\right)\rightarrow ZnCl_2\left(0,04\right)+H_2\left(0,04\right)\)

\(n_{H_2}=0,04\left(mol\right)\)

\(TheoPTHH:n_{Zn}=0,04\left(mol\right)\Rightarrow m_1=2,6\left(g\right)\)

\(n_{HCl}=0,08\left(mol\right)\Rightarrow m_{HCl}=2,92\left(g\right)\)

\(\Rightarrow m_2=\dfrac{2,92.100}{14,6}=20\left(g\right)\)

\(mddsau=2,6+20-0,04.2=22,52(g)\)

\(TheoPTHH:n_{ZnCl_2}=0,04\left(mol\right)\)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,04.136.100}{22,52}=24,26\%\)

1)

a,\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

\(C_{M_{ddNaOH}}=\dfrac{0,2}{0,242}=0,83M\)

\(C\%_{ddNaOH}=\dfrac{8.100\%}{242}=3,3\%\)

b,\(n_{H_2SO_4}=0,1.0,15=0,015\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Mol:      0,03            0,015

\(C_{M_{ddNaOH}}=\dfrac{0,03}{0,2}=0,15M\)

a)

$m_{H_2O} = 242(gam)$

Suy ra : $m_{dd} = 8 + 242 = 250(gam)$

$C\%_{NaOH} = \dfrac{8}{250}.100\% = 3,2\%$

b)

$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} = 2n_{H_2SO_4} = 0,03(mol)$
$C_{M_{NaOH}} = \dfrac{0,03}{0,2} = 0,15M$

a)

$m_{dd} = 16 + 234 = 250(gam)$
$V_{dd} = \dfrac{250}{1,05} = 238(ml)$

$n_{NaOH} = \dfrac{16}{40} = 0,4(mol)$

Suy ra :

$C\%_{NaOH} = \dfrac{16}{250} = 6,4\%$

$C_{M_{NaOH}} = \dfrac{0,4}{0,238} = 1,68M$

b)

$C\%_{NaOH} = \dfrac{16+10}{250+10}.100\% = 10\%$

Câu 1:

Sửa đề: 250ml NaCl 2 mol/l

Ta có: \(n_{NaCl}=0,25\cdot2=0,5\left(mol\right)\) \(\Rightarrow C_{M_{NaCl\left(sau\right)}}=\dfrac{0,5}{0,15+0,25}=1,25\left(M\right)\)

Câu 2:

Ta có: \(n_{NaOH}=0,4\cdot0,25=0,1\left(mol\right)\) 

\(\Rightarrow V_{ddNaOH\left(sau\right)}=\dfrac{0,1}{0,1}=1\left(l\right)\) \(\Rightarrow V_{H_2O\left(thêm\right)}=1-0,4=0,6\left(l\right)\)

Ta có : n_Na \(=\dfrac{4,6}{23}=0,2\left(mol\right)\)

PTHH : \(2Na+2H_2O\rightarrow2NaOH+H_2\uparrow\)

a) Theo pt :\(n_{H_2}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b) Theo pt : \(n_{NaOH}=n_{Na}=0,2\left(mol\right)\)

\(\Rightarrow m_{NaOH}=0,2.40=8\left(g\right)\)

\(m_{ddspu}=4,6+100-0,1.2=104,4\left(g\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{8}{104,4}.100\%=7,66\%\)

(n_{CuSO_4}=dfrac{8}{160}=0,05left(mol ight))

(V_{CuSO_4}=dfrac{m_{CuSO_4}}{D_{CuSO_4}}=dfrac{8}{1,26}=dfrac{400}{63}left(ml ight))

(m_{H_2O}=1.100=100left(g ight))

(n_{H_2O}=dfrac{100}{18}=dfrac{50}{9}left(mol ight))

(C_M=dfrac{0,05+dfrac{50}{9}}{dfrac{400}{63}+100}simeq0,05M)

(C_{\%}=dfrac{8}{100}.100\%=8\%)

Chết làm sai chỗ C_M rồi

Để sửa lại nha, các bác đừng vội ném đá

khối lượng dung dịch là

mdd=mct+mdm=8+100=108(g)

nồng độ phần trăm của dung dịch là

C%=\(\dfrac{mct}{mdd}\).100%

C%=\(\dfrac{8}{108}\).100=7,4%

theo đầu bài ta có khối lượng riêng của dung dịch là

Ddd=D_CuSO4+D_H2O=1,26+1=2,26 g/ml

ta có :Vdd=\(\dfrac{mdd}{Ddd}\)

==>Vdd=\(\dfrac{108}{2,26}\)=47,8ml

n_CuSO4=\(\dfrac{8}{160}\)₌0,05(mol)

nồng độ mol của dung dịch là

C_M=\(\dfrac{n}{V}\)

C_M=\(\dfrac{0,05}{0,0478}\)=1,06M

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)

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