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30 tháng 9 2018

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: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,1\times24=2,4\left(g\right)\)

\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)

\(\Rightarrow\%Mg=\dfrac{2,4}{4,4}\times100\%=54,55\%\)

\(\%MgO=\dfrac{2}{4,4}\times100\%=45,45\%\)

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: \(n_{HCl}=2n_{MgO}=2\times0,05=0,1\left(mol\right)\)

\(\Rightarrow\Sigma n_{HCl}=0,2+0,1=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,3}=1\left(M\right)\)

Theo PT1,2: \(\Sigma n_{MgCl_2}=\dfrac{1}{2}\Sigma n_{HCl}=\dfrac{1}{2}\times0,3=0,15\left(mol\right)\)

\(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,15}{0,3}=0,5\left(M\right)\)

30 tháng 9 2018

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: \(\Sigma n_{MgCl_2}=\dfrac{1}{2}\Sigma n_{HCl}=\dfrac{1}{2}\times0,3=0,15\left(mol\right)\)

\(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,15}{0,3}=0,5\left(M\right)\)

22 tháng 10 2018

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)

27 tháng 9 2021

ko có thể tích khí H2 à

1 tháng 11 2021

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

a. PTHH: 

Mg + 2HCl ---> MgCl2 + H2 (1)

MgO + 2HCl ---> MgCl2 + H2O (2)

Theo PT(1)\(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)

=> \(m_{Mg}=0,1.24=2,4\left(g\right)\)

=> \(m_{MgO}=4,4-2,4=2\left(g\right)\)

b. Ta có: \(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)

=> \(n_{hh}=0,05+0,1=0,15\left(mol\right)\)

Theo PT(1,2)\(n_{HCl}=2.n_{hh}=2.0,15=0,3\left(mol\right)\)

=> \(V_{dd_{HCl}}=\dfrac{0,3}{0,4}=0,75\left(lít\right)\)

30 tháng 9 2018

Mg + 2HCl → MgCl2 + H2 (1)

MgO + 2HCl → MgCl2 + H2O (2)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

a) Theo PT1: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,1\times24=2,4\left(g\right)\)

\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)

\(\Rightarrow\%Mg=\dfrac{2,4}{4,4}\times100\%=54,55\%\)

\(\%MgO=\dfrac{2}{4,4}\times100\%=45,45\%\)

b) 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: \(n_{HCl}=2n_{MgO}=2\times0,05=0,1\left(mol\right)\)

\(\Rightarrow\Sigma n_{HCl}=0,2+0,1=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,3}=1\left(M\right)\)

c) Theo PT1,2: \(\Sigma n_{MgCl_2}=\dfrac{1}{2}\Sigma n_{HCl}=\dfrac{1}{2}\times0,3=0,15\left(mol\right)\)

\(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,15}{0,3}=0,5\left(M\right)\)

12 tháng 12 2021

Đặt \(n_{Mg}=x(mol);n_{Al}=y(mol)\Rightarrow 24x+27y=7,8(1)\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)\\ PTHH:Mg+2HCl\to MgCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow x+1,5y=0,4(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ a,m_{Mg}=0,1.24=2,4(g)\\ m_{Al}=7,8-2,4=5,4(g)\\ b,\Sigma n_{HCl}=2x+3y=0,8(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,8}{2}=0,4(l)\)

\(c,PTHH:MgCl_2+2NaOH\to Mg(OH)_2\downarrow+2NaCl\\ AlCl_3+3NaOH\to Al(OH)_3\downarrow+3NaCl\\ n_{MgCl_2}=n_{Mg(OH)_2}=x=0,1(mol)\\ n_{Al(OH)_3}=n_{AlCl_3}=y=0,2(mol)\\ \Rightarrow m_{Mg(OH)_2}=0,1.58=5,8(g)\\ m_{Al(OH)_3}=0,2.78=15,6(g)\\ m_{kết tủa}=5,8+15,6=21,4(g)\)

nH2=0,1(mol)

PTHH: Mg + 2 HCl -> MgCl2 + H2

0,1__________0,2___________0,1(mol)

MgO + 2 HCl -> MgCl2 + H2O

0,05____0,1___0,05(mol)

mMg=0,1. 24= 2,4(g) -> mMgO=4,4-2,4= 2(g) -> nMgO=0,05((mol)

b) %mMg= (2,4/4,4).100=54,545%

=> %mMgO=45,455%

c) nHCl=0,3(mol) -> mHCl=0,3.36,5=10,95(g)

=> mddHCl=(10,95.100)/7,3=150(g)