a) \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2<----0,4<------0,2<---0,2
\(\%Mg=\dfrac{0,2.24}{14,8}.100\%=32,432\%\)
\(\%MgO=100\%-32,432\%=67,568\%\)
b) \(n_{MgO}=\dfrac{14,8-0,2.24}{40}=0,25\left(mol\right)\)
PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
0,25---->0,5------>0,25
=> \(m_{HCl}=\left(0,5+0,4\right).36,5=32,85\left(g\right)\)
=> \(C\%_{dd.HCl}=\dfrac{32,85}{200}.100\%=16,425\%\)
c)
\(n_{Ba\left(OH\right)_2}=0,25.2=0,5\left(mol\right)\)
PTHH: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
Xét tỉ lệ: \(\dfrac{0,5}{1}>\dfrac{0,9}{2}\) => Ba(OH)2 dư, HCl hết
=> dd thu được có màu hồng
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