\(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=0,2.2=0,4\left(mol\right)\\ C_{MddNaOH}=\dfrac{0,4}{0,4}=1\left(M\right)\)

\(n_{Na_2O}=\dfrac{12,4}{62}=0,2mol\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=0,2.2=0,4mol\\ C_{M_{NaOH}}=\dfrac{0,4}{0,4}=1M\)

Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{Fe}=n_{H_2}=0,25\left(mol\right)\Rightarrow m_{Fe}=0,25.56=14\left(g\right)=m_1\)

\(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{18,25}{7,3\%}=250\left(g\right)=m_2\)

cảm ơn trước nha

1 mol CuO tác dụng với H₂SO₄ thu được 1 mol CuSO₄ 

a, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

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

Ta có: 40n_NaOH + 56n_KOH = 25,44 (1)

\(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}+\dfrac{1}{2}n_{KOH}=0,3.0,9=0,27\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,3\left(mol\right)\\n_{KOH}=0,24\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{NaOH}=0,3.40=12\left(g\right)\\m_{KOH}=0,24.56=13,44\left(g\right)\end{matrix}\right.\)

b, \(m_{ddH_2SO_4}=300.1,14=342\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,27.98}{342}.100\%\approx7,74\%\)

a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

b, \(n_{O_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

Theo PT: \(n_{Al_2O_3}=\dfrac{2}{3}n_{O_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)

c, \(H=\dfrac{18,36}{20,4}.100\%=90\%\)

\(\veebar\)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

a, \(n_{H_2}=\dfrac{0,7437}{24,79}=0,03\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{H_2}=0,06\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{0,06}{0,1}=0,6\left(l\right)\)

b, \(H=\dfrac{669,33}{743,7}.100\%=90\%\)

C16: B

C17: C

C18: B

C19: B

C20: C

\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

\(m_{HCl}=73.20\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

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

Theo PT: \(n_{CuCl_2}=n_{CuO}=0,1\left(mol\right)\)

\(\Rightarrow m_{CuCl_2}=0,1.135=13,5\left(g\right)\)

$n^{CuO}=\genfrac{}{}{0ex}{}{8}{80}=0,1\left(mol\right)$

$m^{HCl}=73.20\%=14,6\left(g\right)\Rightarrow n^{HCl}=\genfrac{}{}{0ex}{}{14,6}{36,5}=0,4\left(mol\right)$

PT: $CuO+2HCl\to CuC{l}^2+H^{2}O$

Xét tỉ lệ: $\genfrac{}{}{0ex}{}{0,1}{1}<\genfrac{}{}{0ex}{}{0,4}{2}$, ta được HCl dư.

Theo PT: $n^{CuC{l}^2}=n^{CuO}=0,1\left(mol\right)$

$\Rightarrow m^{CuC{l}^2}=0,1.135=13,5\left(g\right)$

$Học\; tốt<3$

\(CaCO_3\xrightarrow[]{t^0}CaO+CO_2\)

Cứ 100 tấn đá vôi thì nung được 56 tấn vôi

\(\Rightarrow\dfrac{1}{100}=\dfrac{m_{CaO\left(lt\right)}}{56}\\ \Rightarrow m_{CaO\left(lt\right)}=0,56tấn\\ m_{CaO\left(tt\right)}=0,56.70\%=0,392tấn\)