a) 

$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
$2Zn + O_2 \xrightarrow{t^o} 2ZnO$

b) Theo ĐLBTKL:

\(m_{Kim.lo\text{ại}}+m_{O_2}=m_{\text{ox}it}\)

=> \(m_{O_2}=13,7-11=2,7\left(g\right)\)

Theo ĐLBT KL, có: m_KL + m_O2 = m _oxit

⇒ m_O2 = 13,1 - 1,5 = 11,6 (g)

a)

$2Mg + O_2 \xrightarrow{t^o} 2MgO$
b)

$m_{Mg} + m_{O_2} = m_{MgO}$

c)

$Gọi $m_{Mg} = 3a ; m_{O_2} = 2a$

Ta có:  $3a + 2a = 15 \Rightarrow a = 3$

$m_{O_2} = 3.2 = 6(gam)$

giúp em vs ạ

Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\x_{Ca}=y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}24x+40y=17,6\\x=2y\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,4\\y=0,2\end{matrix}\right.\)

a)\(m_{Mg}=0,4\cdot24=9,6g\)

   \(m_{Ca}=0,2\cdot40=8g\)

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

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

Từ hai pt: \(\Rightarrow\Sigma n_{O_2}=\dfrac{1}{2}n_{Mg}+\dfrac{1}{2}n_{Ca}=\dfrac{1}{2}\cdot0,4+\dfrac{1}{2}\cdot0,2=0,3mol\)

\(\Rightarrow m_{O_2}=0,3\cdot32=9,6g\)

\(V_{O_2}=0,3\cdot22,4=6,72l\)

\(\Rightarrow V_{kk}=5V_{O_2}=5\cdot6,72=33,6l\)

a) 

Có \(\left\{{}\begin{matrix}24.n_{Mg}+40.n_{Ca}=17,6\\n_{Mg}=2.n_{Ca}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}n_{Ca}=0,2\left(mol\right)\\n_{Mg}=0,4\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{Ca}=0,2.40=8\left(g\right)\\m_{Mg}=0,4.24=9,6\left(g\right)\end{matrix}\right.\)

b)

PTHH: 2Ca + O₂ --t^o--> 2CaO 

            0,2-->0,1

             2Mg + O₂ --t^o--> 2MgO

            0,4--->0,2

=> \(V_{O_2}=\left(0,1+0,2\right).22,4=6,72\left(l\right)\)

\(V_{kk}=6,72.5=33,6\left(l\right)\)

Theo ĐLBT KL, có: m_KL + m_O2 = m _oxit

⇒ m_O2 = 28,4 - 15,6 = 12,8 (g)

\(\Rightarrow n_{O_2}=\dfrac{12,8}{32}=0,4\left(mol\right)\Rightarrow V_{O_2}=0,4.22,4=8,96\left(l\right)\)

a) Gọi số mol Cu, Mg là a, b (mol)

=> 64a + 24b = 7,6 (1)

PTHH: 2Cu + O₂ --t^o--> 2CuO

              a---------------->a

           2Mg + O₂ --t^o--> 2MgO

             b------------------>b

=> \(\%MgO=\dfrac{40b}{80a+40b}.100\%=20\%\)

=> a = 2b (2)

(1)(2) => a = 0,1; b = 0,05 

=> \(\left\{{}\begin{matrix}m_{Cu}=0,1.64=6,4\left(g\right)\\m_{Mg}=0,05.24=1,2\left(g\right)\end{matrix}\right.\)

b) 

PTHH: CuO + H₂SO₄ --> CuSO₄ + H₂O 

             0,1--->0,1

            MgO + H₂SO₄ --> MgSO₄ + H₂O 

           0,05--->0,05

=> \(V_{ddH_2SO_4}=\dfrac{0,1+0,05}{0,5}=0,3\left(l\right)\)

\(a,Đặt:n_{MgO}=n_{Mg}=a\left(mol\right)\left(a>0\right)\Rightarrow m_{MgO}=40a\left(g\right)\\ \Rightarrow m_{CuO}=40a.4=160a\left(g\right)\\ n_{CuO}=\dfrac{160}{80}a=2a\left(mol\right)\\ m_{Cu,Mg}=7,6\left(g\right)\\ \Leftrightarrow64.2a+24a=7,6\\ \Leftrightarrow a=0,05\left(mol\right)\\ \Rightarrow m_{Mg}=24.0,05=1,2\left(g\right);m_{Cu}=64.2a=128.0,05=6,4\left(g\right)\\ b,MgO+H_2SO_4\rightarrow MgSO_4+H_2O\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{H_2SO_4}=n_{MgO}+n_{CuO}=a+2a=3a=3.0,05=0,15\left(mol\right)\\ \Rightarrow V_{ddH_2SO_4}=\dfrac{0,15}{0,5}=0,3\left(l\right)\)

a)

2Mg + O₂ --t^o--> 2MgO

2Zn + O₂ --t^o--> 2ZnO

b)

Gọi số mol Mg, Zn là a, b (mol)

=> 24a + 65b = 23,3 (1)

PTHH: 2Mg + O₂ --t^o--> 2MgO

               a-->0,5a------>a

            2Zn + O₂ --t^o--> 2ZnO

               b-->0,5b------>b

=> 40a + 81b = 36,1 (2)

(1)(2) => a = 0,7 (mol); b = 0,1 (mol)

\(n_{O_2}=0,5a+0,5b=0,4\left(mol\right)\)

=> \(V_{O_2}=0,4.22,4=8,96\left(l\right)\)

c) 

m_Mg = 0,7.24 = 16,8 (g)

m_Zn = 0,1.65 = 6,5 (g)

nMgO = 12/40 = 0.3 mol 

Mg + 1/2O2 -to-> MgO 

0.3_____0.15____0.3 

mMg = 0.3*24=7.2 g 

mO2= 0.15*32=4.8 g