Mg + 2HCl → MgCl2 + H2 (1)
MgO + 2HCl → MgCl2 + H2O (2)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT1: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,1\times24=2,4\left(mol\right)\)
\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)
\(\Rightarrow\%Mg=\dfrac{2,4}{4,4}\times100\%=54,55\%\)
\(\%MgO=\dfrac{2}{4,4}\times100\%=45,45\%\)
Theo PT1: \(n_{HCl}=2n_{Mg}=2\times0,1=0,2\left(mol\right)\)
\(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
Theo PT2: \(n_{HCl}=2n_{MgO}=2\times0,05=0,1\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,1+0,2=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,3}=1\left(M\right)\)
Theo PT1,2: \(\Sigma n_{MgCl_2}=\dfrac{1}{2}\Sigma n_{HCl}=\dfrac{1}{2}\times0,3=0,15\left(mol\right)\)
\(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,15}{0,3}=0,5\left(M\right)\)
Mg + 2HCl → MgCl2 + H2 (1)
MgO + 2HCl → MgCl2 + H2O (2)
nH2=2,2422,4=0,1(mol)nH2=2,2422,4=0,1(mol)
Theo PT1: nMg=nH2=0,1(mol)nMg=nH2=0,1(mol)
⇒mMg=0,1×24=2,4(mol)⇒mMg=0,1×24=2,4(mol)
⇒mMgO=4,4−2,4=2(g)⇒mMgO=4,4−2,4=2(g)
⇒%Mg=2,44,4×100%=54,55%⇒%Mg=2,44,4×100%=54,55%
%MgO=24,4×100%=45,45%%MgO=24,4×100%=45,45%
Theo PT1: nHCl=2nMg=2×0,1=0,2(mol)nHCl=2nMg=2×0,1=0,2(mol)
nMgO=240=0,05(mol)nMgO=240=0,05(mol)
Theo PT2: nHCl=2nMgO=2×0,05=0,1(mol)nHCl=2nMgO=2×0,05=0,1(mol)
⇒ΣnHCl=0,1+0,2=0,3(mol)⇒ΣnHCl=0,1+0,2=0,3(mol)
⇒CMHCl=0,30,3=1(M)⇒CMHCl=0,30,3=1(M)
Theo PT1,2: ΣnMgCl2=12ΣnHCl=12×0,3=0,15(mol)ΣnMgCl2=12ΣnHCl=12×0,3=0,15(mol)
⇒CMMgCl2=0,150,3=0,5(M)