a)MgO + 2HCl----->MgCl2 +H2O
b) Ta có
n\(_{MgO}=\frac{4}{40}=0,1\left(mol\right)\)
m\(_{HCl}=\frac{200.14,6}{100}=29,2\left(g\right)\)
n\(_{HCl}=\frac{29,2}{36,5}=0,8\left(mol\right)\)
=> HCl dư
Theo pthh
n\(_{MgCl2}=n_{MgO}=0,1\left(mol\right)\)
m\(_{MgCl2}=0,1.95=9,5\left(g\right)\)
c) Theo pthh
n\(_{HCl}=2n_{MgO}=0,2\left(mol\right)\)
n\(_{HCl}dư=0,8-0,2=0,4\left(mol\right)\)
C%HCl dư =\(\frac{0,4.36,5}{200+4}.100\%=7,16\%\)
C%MgCl2=\(\frac{9,5}{204}.100\%=4,66\%\)
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