Đáp án:

\(\begin{array}{l} a,\ \%m_{Mg}=40\%\\ \%m_{MgO}=60\%\\ b,\ V_{\text{dd HCl}}=63,05\ ml.\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a,\\ PTHH:\\ Mg+2HCl\to MgCl_2+H_2↑\ (1)\\ MgO+2HCl\to MgCl_2+H_2O\ (2)\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\ mol.\\ Theo\ pt\ (1):\ n_{Mg}=n_{H_2}=0,1\ mol.\\ \Rightarrow m_{Mg}=0,1\times 24=2,4\ g.\\ \Rightarrow m_{MgO}=6-2,4=3,6\ g.\\ \Rightarrow \%m_{Mg}=\dfrac{2,4}{6}\times 100\%=40\%\\ \%m_{MgO}=\dfrac{3,6}{6}\times 100\%=60\%\\ b,\\ n_{MgO}=\dfrac{3,6}{40}=0,09\ mol.\\ \Rightarrow \sum n_{HCl}=2n_{Mg}+2n_{MgO}=0,2+0,18=0,38\ mol.\\ \Rightarrow m_{HCl}=0,38\times 36,5=13,87\ g.\\ \Rightarrow m_{\text{dd HCl}}=\dfrac{13,87}{20\%}=69,35\ g.\\ \Rightarrow V_{\text{dd HCl}}=\dfrac{69,35}{1,1}=63,05\ ml.\end{array}\)

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Bài giải:

a.

$n_{H_2}=\frac{2,24}{22,4}=0,1(mol)$ 

$Mg+2HCl→MgCl_2+H_2↑$

  0,1         0,2                        ←  0,1               (mol)   

$MgO+2HCl→MgCl_2+H_2O$

  0,09  →  0,18                                                  (mol)

-$m_{Mg}=0,1.24=2,4(g)$

-$m_{MgO}=m_{hh}-m_{Mg}=6-2,4=3,6(g)$

$n_{MgO}=\frac{3,6}{40}=0,09(mol)$ 

-%$m_{Mg}=\frac{2,4}{6}.100$%$=40$% 

-%$m_{MgO}=100$%$-$%$m_{Mg}=100$%$-40$%$=60$%

b. 

$∑n_{HCl}=0,2+0,18=0,38(mol)$

$m_{chất..tan..HCl}=0,38.36,5=18,37(g)$

$⇒m_{dung..dịch..HCl}=\frac{18,37.100}{20}=69,35(g)$ 

Vì $d_{HCl}=1,1(g/ml)$

$⇒V_{HCl}=\frac{69,35}{1,1}=63,045(ml)$