- Gọi a,b,c,d lần lượt là số mol của MgO, CaO, Mg và Ca trong 10,72 gam X.
\(\Rightarrow40a+56b+24c+40d=10,72\)
\(\Rightarrow5a+7b+3c+5d=1,34\left(1\right)\)
\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{3,248}{22,4}=0,145\left(mol\right)\)
\(m_{H_2}=n.M=0,145\times2=0,29\left(g\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\left(1'\right)\)
\(Ca+2HCl\rightarrow CaCl_2+H_2\left(2'\right)\)
Từ (1'), (2') suy ra: \(n_{Mg}+n_{Ca}=n_{H_2}=0,145\)
\(\Rightarrow c+d=0,145\left(2\right)\Rightarrow c=0,145-d\)
\(n_{MgCl_2}=\dfrac{m}{M}=\dfrac{12,35}{95}=0,13\left(mol\right)\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\left(3'\right)\)
Từ (1'), (3') suy ra: \(\left\{{}\begin{matrix}n_{Mg}+n_{MgO}=n_{MgCl_2}=0,13\\n_{HCl\left(1\right)}=2n_{MgCl_2}=0,26\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow a+c=0,13\left(3\right)\)
\(\left(2\right)-\left(3\right)\Rightarrow d-a=0,015\Rightarrow a=d-0,015\)
Thay \(c=0,145-d\) và \(a=d-0,015\) vào (1) ta được:
\(40\left(d-0,015\right)+56b+24\left(0,145-d\right)+40d=10,72\)
\(\Rightarrow\left(40-24+40\right)d+56b+3,48-0,6=10,72\)
\(\Rightarrow56\left(b+d\right)=7,84\)
\(\Rightarrow b+d=0,14\)
\(CaO+2HCl\rightarrow CaCl_2+H_2O\left(4'\right)\)
Từ (2'), (4') \(\Rightarrow n_{HCl\left(2\right)}=2\left(n_{Ca}+n_{CaO}\right)=2\left(d+b\right)=0,28\left(mol\right)\)
\(n_{HCl}=n_{HCl\left(1\right)}+n_{HCl\left(2\right)}=0,26+0,28=0,54\left(mol\right)\)
Từ (1'), (2') \(\Rightarrow n_{HCl\left(3\right)}=2n_{H_2}=...=0,29\left(mol\right)\)
\(\Rightarrow n_{HCl\left(4\right)}=n_{HCl}-n_{HCl\left(3\right)}=...=0,25\left(mol\right)\)
Từ (3'), (4') \(\Rightarrow n_{H_2O}=\dfrac{n_{HCl\left(4\right)}}{2}=...=0,125\left(mol\right)\)
\(\left\{{}\begin{matrix}m_{HCl}=n.M=0,54\times36,5=19,71\left(g\right)\\m_{H_2O}=n.M=0,125\times18=2,25\left(g\right)\end{matrix}\right.\)
Theo định luật bảo toàn KL ta có:
\(m_X+m_{HCl}=\left(m_{MgCl_2}+m_{CaCl_2}\right)+\left(m_{H_2}+m_{H_2O}\right)\)
\(\Rightarrow x=15,54\left(g\right)\)
