a) Mg + 2HCl → MgCl2 + H2 (1)
MgO + 2HCl → MgCl2 + H2O (2)
b) \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,2\times2=0,4\left(g\right)\)
Theo PT1: \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,2\times24=4,8\left(g\right)\)
\(\Rightarrow m_{MgO}=8,8-4,8=4\left(g\right)\)
\(\Rightarrow\%m_{Mg}=\dfrac{4,8}{8,8}\times100\%=54,55\%\)
\(\%m_{MgO}=\dfrac{4}{8,8}\times100\%=45,45\%\)
c) \(m_{dd}saupư=8,8+400-0,4=408,4\left(g\right)\)
\(m_{HCl}=400\times7,3\%=29,2\left(g\right)\)
Theo PT1: \(n_{HCl}=2n_{Mg}=2\times0,2=0,4\left(mol\right)\)
\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)
Theo PT2: \(n_{HCl}=2n_{MgO}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}pư=0,4+0,2=0,6\left(mol\right)\)
\(\Rightarrow m_{HCl}pư=0,6\times36,5=21,9\left(g\right)\)
\(\Rightarrow m_{HCl}dư=29,2-21,9=7,3\left(g\right)\)
\(\Rightarrow C\%_{HCl}dư=\dfrac{7,3}{408,4}\times100\%=1,79\%\)
Theo PT1: \(n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\)
Theo Pt2: \(n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\)
\(\Rightarrow\Sigma n_{MgCl_2}=0,2+0,1=0,3\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=0,3\times95=28,5\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{28,5}{408,4}\times100\%=6,98\%\)
a. nHCl=\(\dfrac{400\times7.3}{100\times36.5}\)= 0,8 (mol) ; nH2=\(\dfrac{4,48}{22,4}\)=0,2 (mol)
Mg + 2HCl -> MgCl2 + H2
MgO + 2HCl -> MgCl2 + H2O
b. nH2= nMg= nMgCl2 0,2 mol
mMg= 24\(\times\)0,2= 4,8 (g) \(\Rightarrow\) %mMg= \(\dfrac{4,8\times100}{8,8}\)\(\approx\) 54,5%
\(\rightarrow\)nMgCl2= nMgO= \(\dfrac{8.8-4.8}{40}\)= 0,1 (mol)
\(\rightarrow\) %mMgO= 100 - 54,5 = 45,5%
c. mdd sau pư= 8,8 + 400 - 0.2\(\times\)2 = 408,4 (g)
nMgCl2sau pư= 0,2 + 0,1 = 0,3 (mol)
C%MgCl2=\(\dfrac{95\times0,3\times100}{408,4}\)\(\approx\)7%
