a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

b, Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{MgO}=y\left(mol\right)\end{matrix}\right.\)

⇒ 24x + 40y = 4,4 (1)

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

Theo PT: \(n_{H_2}=n_{Mg}=x\left(mol\right)\)

⇒ x = 0,1 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1.24}{4,4}.100\%\approx54,54\%\\\%m_{MgO}\approx45,46\%\end{matrix}\right.\)

c, Theo PT: \(\Sigma n_{HCl}=2n_{Mg}+2n_{MgO}=0,3\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{0,3}{2}=0,15\left(l\right)=150\left(ml\right)\)

Bạn tham khảo nhé!

1)

$MgO + 2HCl to MgCl_2 + H_2O$
$Mg + 2HCl \to MgCl_2 + H_2$

2)

$n_{Mg} = n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)$
$m_{Mg} = 0,1.24 = 2,4(gam)$
$m_{MgO} = 4,4 - 2,4 = 2(gam)$

3)

$n_{HCl} = 2n_{Mg} + 2n_{MgO} = 0,1.2 + \dfrac{2}{40}.2 = 0,3(mol)$
$V_{dd\ HCl} = \dfrac{0,3}{2} = 0,15(lít) = 150(ml)$

\(a.Mg+2HCl\rightarrow MgCl_2+H_2\\ MgO+2HCl\rightarrow MgCl_2+H_2O\\ b.n_{H_2}=n_{Mg}=0,1\left(mol\right)\\ \Rightarrow m_{Mg}=2,4\left(g\right)\\ \Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\\ c.\%m_{Mg}=\dfrac{2,4}{4,4}.100=54,55\%\\ \%m_{MgO}=45,45\%\\ d.\Sigma n_{HCl}=2n_{H_2}+2n_{MgO}=0,1.2+\dfrac{2}{40}.2=0,3\left(mol\right)\\ CM_{HCl}=\dfrac{0,3}{2}=0,15\left(l\right)=150ml\)

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)

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

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

a) nH2=0,2(mol)

PTHH: Mg + 2 HCl -> MgCl2 + H2

0,2______0,4_______0,2_____0,2(mol)

PTHH: MgO +2 HCl -> MgCl2 + H2O

b) mMgO= 8,8 - 0,2.24=4(g)

%mMgO= (4/8,8).100= 45,455%

=>%Mg=54,545%

c) nHCl(tổng)= 2. nMg + 2. nMgO= 2. 0,2+ 0,1.2=0,6(mol)

=> mHCl= 0,6.36,5=21,9(g)

=>mddHCl=(21,9.100)/7,3=300(g)

d) mddMgCl2= mddHCl + m(hỗn hợp ban đầu) - mH2

<=>mddHCl= 300+ 8,8- 0,2.2= 308,4(g)

nMgCl2=0,3(mol) => mMgCl2= 0,3.95=28,5(g)

=>C%ddMgCl2= (28,5/308,4).100=9,241%