1. \(CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl_{\downarrow}\)

\(n_{AgCl}=\dfrac{28,7}{143,5}=0,2\left(mol\right)\)

Theo PT: \(n_{Ca\left(NO_3\right)_2}=\dfrac{1}{2}n_{AgCl}=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{Ca\left(NO_3\right)_2}}=\dfrac{0,1}{0,15+0,05}=0,5\left(M\right)\)

2. \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)

BTNT Na, có: n_NaOH = n_Na = 0,1 (mol)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(2Al+2NaOH+2H_2O\rightarrow2NaAlO_2+3H_2\)

Xét tỉ lệ \(\dfrac{0,2}{2}>\dfrac{0,1}{2}\), ta được Al dư.

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{NaOH}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)\)

\(n_{Al\left(pư\right)}=n_{NaOH}=0,1\left(mol\right)\Rightarrow n_{Al\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

\(\Rightarrow m_{Al\left(dư\right)}=0,1.27=2,7\left(g\right)\)

Câu 1
\(a)PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b)200ml=0,2l\\ n_{HCl}=0,2.1=0,2mol\\ n_{H_2}=n_{MgCl_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}\cdot0,2=0,1mol\\ V_{H_2}=0,1.24,79=2,479l\\ c)C_{M_{MgCl_2}}=\dfrac{0,1}{0,2}=0,5M\)

Câu 2
\(a)PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b)n_{Mg}=\dfrac{1,2}{24}=0,05mol\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,05mol\\ V_{H_2}=0,05.24,79=1,2395l\\ c)n_{HCl}=2n_{Mg}=2.0,05=0,1mol\\ V_{ddHCl}=\dfrac{0,1}{2}=0,05l\\ d)C_{M_{MgCl_2}}=\dfrac{0,05}{0,05}=1M\)

a) $2NaOH + Cl_2 \to NaCl + NaCl + H_2O$

b) $n_{Cl_2} = \dfrac{6,5}{22,4} = 0,29(mol)$

$n_{NaOH} = 2n_{Cl_2} = 0,58(mol)$

$V_{dd\ NaOH} = \dfrac{0,58}{1} = 0,58(lít)$

$n_{NaCl} = n_{NaClO} = n_{Cl_2} = 0,29(mol)$
$C_{M_{NaCl}} = C_{M_{NaClO}} = \dfrac{0,29}{0,58} = 0,5M$

a, \(n_{Cu}=\dfrac{9,6}{64}=0,15\left(mol\right)\)

PTHH: 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O

Mol:     0,15                          0,15            0,1

\(V_{NO}=0,1.22,4=2,24\left(l\right)\)

b, \(C_{M_{ddCu\left(NO_3\right)_2}}=\dfrac{0,15}{0,2}=0,75M\)

\(n_{CO_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\)

PTHH :

\(CO_2+2KOH\rightarrow K_2CO_3+H_2O\)

0,125    0,25          0,125 

\(a,C_{M\left(KOH\right)}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)

\(b,C_{M\left(K_2CO_3\right)}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)

Cứu mình với

Đề bài có vấn đề

\(n_{Cl_2}=\dfrac{3,36}{22,4}=0,15mol\\ Cl_2+2NaOH\rightarrow NaCl+NaOH+H_2O\\ n_{NaCl}=n_{NaOH}=n_{Cl_2}=0,15mol\\ C_{M_{NaCl}}=\dfrac{0,15}{0,3}=0,5M\\ C_{M_{NaClO}}=\dfrac{0,15}{0,3}=0,5M\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2mol\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

0,2     0,2                0,2       0,2

a)\(V_{H_2}=0,2\cdot22,4=4,48l\)

b)\(C_{M_{H_2SO_4}}=\dfrac{0,2}{0,5}=0,4M\)

c)\(C_{M_{FeSO_4}}=\dfrac{0,2}{0,5}=0,4M\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\) 
pthh : \(Fe+H_2SO_4->FeSO_4+H_2\) 
          0,2                                      0,2
=> \(V_{H_2}=0,2.22,4=4,48\left(L\right)\) 
\(m_{H_2SO_4}=\dfrac{0,5}{22,4}.98\approx2,188\left(g\right)\) 
=> mdd=11,2+2,188=13,388(g) 
C%=\(\dfrac{2,188}{13,388}.100\%=16,3\%\)