1,a,Gọi \(n_{Al}=a\left(mol\right)\rightarrow n_{Mg}=0,5a\left(mol\right)\)

\(\rightarrow27a+24.0,5b=7,8\\ \Leftrightarrow a=0,2\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Mg}=0,1\left(mol\right)\end{matrix}\right.\)

b, \(\rightarrow\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Mg}=0,1.24=2,4\left(g\right)\end{matrix}\right.\)

2, \(n_{O_2}=\dfrac{0,16}{32}=0,005\left(mol\right)\)

PTHH: 2HgO --t^o--> 2Hg + O₂

                                 0,01<- 0,05

\(\rightarrow m_{Hg}=0,01.201=2,01\left(g\right)\)

Gọi nAl=2a=>nMg=a mol

=>mhh=2a.27+24a=7,8=>a=0,1 mol

Vậy Al 0,2 mol Mg 0,1 mol

=>mAl=0,2.27=5,4gam

mMg=0,1.24=2,4gam

đề hình như thiếu á

a) %nMg = (1:3).100% = 33.33%
%nAl = (2:3).100% = 66.67%
b) gọi nMg là x (mol) --> nAl là 2x (mol)
Ta có pt: 27.2x + 24x = 7.8 --> x=0.1
---> nAl=0.2mol; nMg=0.1mol
---> mAl=5.4g; mMg=2.4g

Gọi a là số mol Mg

Ta có

\(n_{Al}:n_{Mg}=2:1\Rightarrow n_{Al}=2a\left(mol\right)\)

\(27.2a+24a=7,8\)

\(\Rightarrow a=0,1\left(mol\right)\)

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

\(n_{AL}=0,2\left(mol\right)\)

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

\(m_{Al}=0,2.27=5,4\left(g\right)\)

a, Ta có : \(\dfrac{n_{Al}}{n_{Mg}}=\dfrac{2}{1}\)

Mà \(m_{hh}=m_{Al}+m_{Mg}=27n_{Al}+24n_{Mg}=7,8\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,2\\n_{Mg}=0,1\end{matrix}\right.\) mol

b, Ta có : \(\left\{{}\begin{matrix}m_{Al}=n.M=5,4\\m_{Mg}=n.M=2,4\end{matrix}\right.\) g

Vậy ...

a)Gọi $n_{Al} =2 a(mol) ; n_{Zn} = a(mol) \Rightarrow 2a.27 + 65a = 23,8$

$\Rightarrow a = 0,2(mol)$

Suy ra : $n_{Al} = 0,4(mol) ; n_{Zn} = 0,2(mol)$

b) $m_{Al} = 0,4.27 = 10,8(gam) ; m_{Zn} = 0,2.65 = 13(gam)$

a) \(n_{MgO}=\dfrac{16}{40}=0,4\left(mol\right)\)

=> n_Mg = 0,4 (mol)

=> m_Mg = 0,4.24 = 9,6 (g)

b) n_Mg = 0,4 (mol) => n_X = 0,6 (mol)

m_X = 33,6 - 9,6 = 24 (g)

=> \(M_X=\dfrac{24}{0,6}=40\left(g/mol\right)\)

=> X là Ca

c) 

PTHH: 2Mg + O₂ --t^o--> 2MgO

             2Ca + O₂ --t^o--> 2CaO

\(m_{O_2}=49,6-33,6=16\left(g\right)\)

=> \(n_{O_2}=\dfrac{16}{32}=0,5\left(mol\right)\)

=> V_O2 = 0,5.22,4 = 11,2 (l)

=> V_kk = 11,2.5 = 56 (l)

a)\(n_{MgO}=\dfrac{16}{40}=0,4mol\Rightarrow n_{Mg}=0,4mol\Rightarrow m_{Mg}=9,6g\)

\(\Rightarrow m_X=33,6-9,6=24g\)

b)Theo bài: \(\dfrac{n_{Mg}}{n_X}=\dfrac{2}{3}\Rightarrow\dfrac{0,4}{n_X}=\dfrac{2}{3}\Rightarrow n_X=0,6mol\)

\(\Rightarrow M_X=\dfrac{24}{0,6}=40\Rightarrow X:Ca\)

c)\(2Mg+O_2\underrightarrow{t^o}2MgO\)

     \(x\)                    \(x\)

   \(2Ca+O_2\underrightarrow{t^o}2CaO\)

    \(y\)                      \(y\)

\(\Rightarrow\left\{{}\begin{matrix}24x+40y=33,6\\40x+56y=49,6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,4\\y=0,6\end{matrix}\right.\)

\(\Sigma n_{O_2}=\dfrac{1}{2}n_{Mg}+\dfrac{1}{2}n_{Ca}=0,2+0,3=0,5mol\)

\(m_{O_2}=0,5\cdot32=16g\)

\(V_{kk}=5V_{O_2}=5\cdot0,5\cdot22,4=56l\)

a) \(n_{MgO}=\dfrac{16}{40}=0,4\left(mol\right)\)

=> n_Mg = 0,4 (mol)

=> m_Mg = 0,4.24 = 9,6 (g)

b)

Có: n_Mg : n_X = 2 : 3

Mà n_Mg = 0,4 (mol)

=> n_X = 0,6 (mol)

m_X = 33,6 - 9,6 = 24 (g)

=> \(M_X=\dfrac{24}{0,6}=40\left(g/mol\right)\)

=> X là Ca

c)

PTHH: 2Mg + O₂ --t^o--> 2MgO

            0,4->0,2

             2Ca + O₂ --t^o--> 2CaO

              0,6->0,3

=> \(m_{O_2}=\left(0,2+0,3\right).32=16\left(g\right)\)

\(V_{O_2}=\left(0,2+0,3\right).22,4=11,2\left(l\right)\)

=> V_kk = 11,2.5 = 56 (l)

a) MX=240,6=40(g/mol)MX=240,6=40(g/mol)

=> X là Ca

c PTHH: 2Mg + O₂ --t^o--> 2MgO

            0,4->0,2

             2Ca + O₂ --t^o--> 2CaO

              0,6->0,3

=> mO2=(0,2+0,3).32=16(g)mO2=(0,2+0,3).32=16(g)

VO2=(0,2+0,3).22,4=11,2(l)VO2=(0,2+0,3).22,4=11,2(l)

=> V_kk = 11,2.5 = 56 (l)