Hóa học

hời ơiiii lm onn cíuuu giúp linh hồn này, mai tui đi thi gòiiii

\(n_{C_2H_4}=\dfrac{2,8}{28}=0,1mol\\ \%V_{C_2H_4}=\dfrac{0,1.22,4}{11,2}\cdot100\%=20\%\\ \%V_{CH_4}=100\%-20\%=80\%\)

\(n_{H_2}=\dfrac{9,916}{24,79}=0,4mol\\ n_{Zn}=a;n_{Al}=b\\ Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\\ 2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}a+1,5b=0,4\\65a+27b=11,9\end{matrix}\right.\\ \Rightarrow a=0,1;b=0,2\\ \%m_{Zn}=\dfrac{0,1.65}{11,9}\cdot100\%=54,62\%\\ \%m_{Al}=100\%-54,62\%=45,38\%\)

\(m_{HCl}=\dfrac{200\cdot3,65\%}{100\%}=7,3\left(g\right)\)

\(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)

PTHH: \(2HCl+Ba\left(OH\right)_2\rightarrow BaCl_2+2H_2O\)

Mol(đề):  2                1                    1              2

Mol(pt);  0,2

Theo pt, ta có: \(n_{Ba\left(OH\right)_2}=\dfrac{n_{HCl}}{2}=\dfrac{0,2}{2}=0,1\left(mol\right)\)

\(m_{Ba\left(OH\right)_2}=0,1\cdot171=17,1\left(g\right)\)

\(\Rightarrow m_{\text{dd }Ba\left(OH\right)_2}=\dfrac{17,1\cdot100\%}{1,71\%}=1000\left(g\right)\)

B cùng có hóa trị II

\(a.n_{H_2}=\dfrac{3,7185}{24,79}=0,15mol\\ Mg+2HCl\rightarrow MgCl_2+H_2\)

0,15        0,3                0,15              0,15

\(\%m_{Mg}=\dfrac{0,15.24}{96}\cdot100\%=3,75\%\\ \%m_{MgO}=96,25\%\\ b.n_{MgO}=\dfrac{96.96,25\%}{40}=2,31mol\\ MgO+2HCl\rightarrow MgCl_2+H_2O\)

2,31           4,62              2,31                  2,31

\(C_{\%HCl}=\dfrac{\left(0,3+4,62\right).36,5}{150}\cdot100\%=119,72\%?\\ c.C_{\%MgCl_2}=\dfrac{\left(0,15+2,31\right).95}{96+150-0,15.2}\cdot100\%=95,12\%\)