C₂H₂ + \(\frac{5}{2}\)O₂ => (t^o) 2CO₂ + H₂O

0.25 0.625 0.5 0.25 (mol)

n_CO2 = V/22.4 = 11.2/22.4 = 0.5 (mol);

m₁ = n.M = 0.25 x 26 = 6.5 (g);

m₂ = n.M = 0.25 x 18 = 4.5 (g);

m₁ + m₂ = 6.5 + 4.5 = 11 (g);

V = VO2 = n.22.4 = 0.625x22.4=14(l)

nZn = 13/65 = 0.2 (mol) 

2Zn + O2 -to-> 2ZnO 

0.2......0.1

VO2 = 0.1 * 22.4 = 2.24 (l)

PTHH: \(Zn+\dfrac{1}{2}O_2\xrightarrow[]{t^o}ZnO\)

Bảo toàn khối lượng: \(m_{O_2}=m_{ZnO}-m_{Zn}=1,6\left(g\right)\)

a. \(2Zn+O_2\rightarrow2ZnO\)

b.\(m_{Zn}+m_{O_2}\rightarrow m_{ZnO}\)

\(\Rightarrow6,5+m_{O_2}=8,1\)

\(\Rightarrow m_O=8,1-6,5=1,6\)

nZn = 13/65 = 0,2 (mol)

nO2 = 4,48/22,4 = 0,2 (mol)

PTHH: 2Zn + O2 -> (t°) 2ZnO

LTL: 0,2/2 < 0,2 => O2 dư

nO2 (p/ư) = 0,2/2 = 0,1 (mol)

mO2 (dư) = (0,2 - 0,1) . 32 = 3,2 (g)

nZnO = nZn = 0,2 (mol)

mZnO = 0,2 . 81 = 16,2 (g)

Dòng thứ 2
No2 = 4,48/22,4=0,2(mol)
Tại sao lại chia cho 22,4
Mà trong công thức là
N=m/M thì sẽ là 
4,48/16 ( 16 = nguyên tử khối oxi)

nCu=0,3mol

pthh: 2Cu+O2=> 2CuO

            0,3->0,15->0,3

=> m1=0,3.80=24g

=> v=0,15.22,4=3,36l

CuO+2HCl=>CuCl2+H2O

0,3->0,6

=> m2=0,6.36,5=21,9g

nCuO = 19,2 : 64 = 0,3 mol

PTHH:      2Cu + O2 ===> 2CuO

                   0,3     0,15          0,3                  (mol)

                  CuO + 2HCl ===> CuCl2 + H2O

                    0,3         0,6                               (mol)

Lập tỉ lệ các số mol theo pt, ta có:

               V  =  0,15 x 22,4 = 3,36 lít

               m1 = 0,3 x 80 = 24 gam

               m2 = 0,6 x 36,5 = 21,9 gam

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ PTHH:4Al+3O_2\rightarrow^{t^o}2Al_2O_3\\ \Rightarrow\left\{{}\begin{matrix}n_{O_2}=\dfrac{3}{4}n_{Al}=0,3\left(mol\right)\\n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}V=V_{O_2}=0,3\cdot22,4=6,72\left(l\right)\\a=m_{Al_2O_3}=0,2\cdot102=20,4\left(g\right)\end{matrix}\right.\)

\(Ta.có:\dfrac{n_{Mg}}{n_{Zn}}=\dfrac{3}{1}\Rightarrow\dfrac{\dfrac{m_{Mg}}{M_{Mg}}}{\dfrac{m_{Zn}}{M_{Zn}}}=\dfrac{m_{Mg}}{24}.\dfrac{65}{m_{Zn}}=\dfrac{3}{1}\Rightarrow\dfrac{m_{Mg}}{m_{Zn}}=\dfrac{3}{1}:\dfrac{65}{24}=\dfrac{72}{24}\)

\(\Rightarrow n_{Mg}=13,7:\left(72+24\right).3=0,3\left(mol\right)\\ \Rightarrow n_{Zn}=\dfrac{n_{Mg}}{3}=\dfrac{0,3}{3}=0,1\left(mol\right)\)

\(\Rightarrow m_{Mg}=n.M=0,3.24=7,2\left(g\right)\\ \Rightarrow m_{Zn}=n.M=0,12.65=6,5\left(g\right)\)

\(a,PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\left(1\right)\\ PTHH:2Zn+O_2\underrightarrow{t^o}2ZnO\left(2\right)\)

\(Theo.PTHH\left(1\right):n_{O_2\left(1\right)}=\dfrac{1}{2}.n_{Mg}=\dfrac{1}{2}.0,3=0,15\left(mol\right)\\ Theo.PTHH\left(2\right):n_{O_2\left(2\right)}=\dfrac{1}{2}.n_{Zn}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\\ n_{O_2\left(tổng\right)}=n_{O_2\left(1\right)}+n_{O_2\left(2\right)}=0,15+0,05=0,2\left(mol\right)\\ V_{O_2\left(tổng,đktc\right)}=n.22,4=0,2.22,4=4,48\left(l\right)\)

\(b,Theo.PTHH\left(1\right):n_{MgO}=n_{Mg}=0,3\left(mol\right)\\ m_{MgO}=n.M=0,3.40=12\left(g\right)\\ Theo.PTHH\left(2\right):n_{ZnO}=n_{Zn}=0,1\left(mol\right)\\ m_{ZnO}=n.M=0,1.81=8,1\left(g\right)\\ m=m_{hh}=m_{MgO}+m_{ZnO}=12+8,1=20,1\left(g\right)\)