a)

\(n_{Mg} = \dfrac{4,8}{24} = 0,2(mol)\\ n_{O_2} = \dfrac{3,36}{22,4}= 0,15(mol)\\ 2Mg + O_2 \xrightarrow{t^o} 2MgO\)

Ta thấy : 

\( \dfrac{n_{Mg}}{2} = 0,1 < n_{O_2} = 0,15 \) nên O₂ dư.

\(n_{O_2\ pư} = \dfrac{n_{Mg}}{2} = 0,1(mol)\\ m_{O_2\ dư} = (0,15-0,1).32 = 1,6(gam)\\ V_{O_2\ dư} = (0,15-0,1).22,4 = 1,12(lít)\)

b)

\(n_{MgO} = n_{Mg} = 0,2\ mol\\ \Rightarrow m_{MgO} = 0,2.40 = 8\ gam\)

a, Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Xét tỉ lệ: \(\dfrac{0,3}{2}< \dfrac{0,2}{1}\), ta được O₂ dư.

Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=0,15\left(mol\right)\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,05\left(mol\right)\Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)

b, \(n_{H_2O}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{H_2O}=0,3.18=5,4\left(g\right)\)

c, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

_______0,3_______________________0,15 (mol)

\(\Rightarrow m_{KMnO_4}=0,3.158=47,4\left(g\right)\)

Bạn tham khảo nhé!

nP=\(\dfrac{62}{31}\)=0,2(mol)

nO2=\(\dfrac{7,84}{22,4}\)=0,35(mol)

PTHH:4P+5O2to→2P2O5

tpứ:    0,2     0,35     

pứ:     0,2     0,25     0,1

spứ:     0     0,1         0,1

a)chất còn dư là oxi

mO2dư=0,1.32=3,2(g)

b)mP2O5=n.M=0,1.142=14,2(g)

\(a.n_P=0,2\left(mol\right);n_{O_2}=0,35\left(mol\right)\\ 4P+5O_2-^{t^o}\rightarrow2P_2O_5\\ LTL:\dfrac{0,2}{4}< \dfrac{0,35}{5}\\ \Rightarrow SauphảnứngO_2dư\\ n_{O_2\left(pứ\right)}=\dfrac{5}{4}n_P=0,25\left(mol\right)\\ \Rightarrow m_{P\left(dư\right)}=\left(0,35-0,25\right).32=3,2\left(g\right)\\ b.n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\\ \Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)

a) n\(_P\) = \(\dfrac{6,2}{31}\) = 0,2(mol); n\(_{O_2}\) = \(\dfrac{7,84}{22,4}\) = 0,35(mol).

Theo phương trình hóa học:

\(4P+5O_2\) \(\underrightarrow{t^o}\) \(2P_2O_5\)

4        5          2 (mol)

0,2 \(\rightarrow\)0,25 \(\rightarrow\) 0,1(mol)

Tỉ lệ số mol: \(\dfrac{0,2}{4}< \dfrac{0,35}{5}\) \(\Rightarrow\) \(O_2\) phản ứng dư, P phản ứng hết.

\(n_{O_2}\)dư = 0,35 - \(\dfrac{0,2.5}{4}\) = 0,1(mol) \(\Rightarrow\) \(m_{O_2}\)dư = 0,1.32 = 3,2g

b) \(n_{P_2O_5}\) = 0,1.142 = 14,2g

nFe = 16.8/56 = 0.3 (mol) 

nO2 = 6.72/22.4 = 0.3 (mol) 

2Fe + 3O2 -to-> Fe3O4 

0.2___0.3________0.1 

mFe dư = ( 0.3 - 0.2 ) * 56 = 5.6 (g) 

mFe3O4 = 0.1*232 = 23.2 (g) 

 a)

b)

nO2=6,72/22,4=0,3mol

Ta có: 

mFe3O4=0,1.232=23,2g

Bài 1:

\(a,2Cu+O_2\underrightarrow{t^o}2CuO\)

b, \(n_{O_2}=\dfrac{1,12}{32}=0,035mol\)

\(n_{Cu}=\dfrac{6,4}{64}=0,1mol\)

\(\dfrac{0,1}{2}>\dfrac{0,035}{1}\) => Cu dư, O₂ đủ

\(n_{Cu}\left(dư\right)=0,1-0,07=0,039\left(mol\right)\)

c, \(m_{CuO}=0,07.80=5,6g\)

Bài 2:

\(n_{Al}=\dfrac{13,5}{27}=0,5mol\)

\(n_{O_2}=\dfrac{6,67}{32}=0,21\left(mol\right)\)

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

\(\dfrac{0,5}{4}>\dfrac{0,21}{3}\) => Al dư, O₂ đủ

\(n_{Al_2O_3}=\dfrac{2}{3}.0,21=0,14\left(mol\right)\)

\(m_{Al_2O_3}=0,14.102=14,28g\)

Cảm ơn bạn @anayuiky đã nhắc lỗi sai. Mình sửa lại ý c):

PTHH: \(2KMnO_4\rightarrow^{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)

Theo phương trình \(n_{KMnO_4}=n_{O_2}.2=0,25.2=0,5mol\)

\(\rightarrow m_{KMnO_4}=0,5.\left(39+55+16.4\right)=79g\)

a. \(n_{H_2}=\frac{V}{22,4}=\frac{11,2}{22,4}=0,5mol\)

\(n_{O_2}=\frac{V}{22,4}=\frac{10,08}{22,4}=0,45mol\)

PTHH:        \(2H_2+O_2\rightarrow^{t^o}2H_2O\)

Ban đầu:     0,5         0,45                     mol

Trong pứng:  0,5         0,25          0,5     mol

Sau pứng:       0          0,2            0,5      mol

\(\rightarrow M_{O_2\left(dư\right)}=n.M=0,2.32=6,4g\)

b. Theo phương trình \(n_{H_2O}=n_{H_2}=0,5mol\)

\(\rightarrow m_{H_2O}=n.M=0,5.18=9g\)

c. PTHH: \(2KMnO_4\rightarrow^{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)

                      0,9                                                            0,45          mol

\(\rightarrow n_{KMnO_4}=\frac{2}{1}n_{O_2}=\frac{0,45.2}{1}=0,9mol\)

\(\rightarrow m_{KMnO_4}=n.M=0,9.158=142,2g\)

ủa bạn ơi hình như câu c sai sai j á
Mình tưởng là:
PTHH: 2KMnO₄ --t^o--> K₂MnO₄ + MnO₂ + O₂
             0,5<-----------------------------------0,25
=> m_KMnO4  = 0,5.158 = 79(g)

nP = 6.2/31 = 0.2 (mol) 

nO2 = 6.72/22.4 = 0.3 (mol) 

4P + 5O2 -to-> 2P2O5 

0.2___0.25_____0.1

mO2 dư = ( 0.3 - 0.25) * 32 = 1.6(g) 

mP2O5 = 0.1*142 = 14.2 (g) 

Ta có: \(n_P=\dfrac{6.2}{31}=0.29mol\)

\(n_{O_2}=\dfrac{6.72}{22.4}=0.3mol\)

PTHH:

\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

ta có:

\(\left\{{}\begin{matrix}\dfrac{n_{P\left(bra\right)}}{nP_{\left(pthh\right)}}=\dfrac{0.2}{4}=0.05\\\dfrac{n_{O_2\left(bra\right)}}{n_{O_2}\left(pthh\right)}=\dfrac{0.3}{5}=0.06\end{matrix}\right.\)

=> \(O_2\) dư 

\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

4   ----------->2                

0.2---------->0.1=n_P2O5

=>\(m_{P_2O_5}=142.0.1=14.2\left(g\right)\)

\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ LTL:\dfrac{0,2}{4}< \dfrac{0,4}{5}\Rightarrow O_2dư\)

\(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)

\(n_{P_2O_5\left(lt\right)}=\dfrac{1}{2}n_P=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{P_2O_5\left(lt\right)}=0,1.142=14,2\left(g\right)\\ m_{P_2O_5\left(tt\right)}=0,1.142.80\%=11,36\left(g\right)\)

$a) CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$

b) $n_{CH_4} = \dfrac{3,92}{22,4} = 0,175(mol)$

$n_{O_2} = \dfrac{3,84}{32} = 0,12(mol)$

Ta thấy : $n_{CH_4} : 1 > n_{O_2} : 2$ nên $CH_4$ dư

$n_{CH_4\ pư} = \dfrac{1}{2}n_{O_2} = 0,06(mol)$
$\Rightarrow m_{CH_4\ dư} = (0,175 - 0,06).16 = 1,84(gam)$

c) $2NaOH + CO_2 \to Na_2CO_3 + H_2O$

Theo PTHH :

$n_{Na_2CO_3} = n_{CO_2} = \dfrac{1}{2}n_{CH_4} = 0,06(mol)$
$m_{Na_2CO_3} = 0,06.106 = 6,36(gam)$