Bài 2:

\(n_{Zn}=\dfrac{15,6}{65}=0,24\left(mol\right);n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ a,Vì:\dfrac{0,4}{1}>\dfrac{0,24}{1}\Rightarrow H_2SO_4dư\\ n_{H_2\left(LT\right)}=n_{H_2SO_4\left(p.ứ\right)}=n_{Zn}=0,24\left(mol\right)\\ a,n_{H_2\left(TT\right)}=50\%.0,24=0,12\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc,thực.tế\right)}=0,12.22,4=2,688\left(l\right)\\ b,n_{H_2SO_4\left(dư\right)}=0,4-0,24=0,16\left(mol\right)\\ \Rightarrow m_{H_2SO_4\left(dư\right)}=0,16.98=15,68\left(g\right)\)

Bài trên

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right);n_{O_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\\ a,PTHH:2H_2+O_2\rightarrow\left(t^o\right)2H_2O\\ Vì:\dfrac{1}{1}>\dfrac{0,5}{2}\Rightarrow O_2thừa\\ n_{O_2\left(thừa\right)}=1-\dfrac{0,5}{2}=0,75\left(mol\right)\\ \Rightarrow m_{O_2\left(thừa\right)}=0,75.32=24\left(g\right)\\ b,n_{H_2O}=n_{H_2}=0,5\left(mol\right)\Rightarrow m_{H_2O}=0,5.18=9\left(g\right)\)

\(a,n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right);n_{O_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\\ a,PTHH:4Na+O_2\rightarrow2Na_2O\\ Vì:\dfrac{0,1}{4}< \dfrac{0,04}{1}\Rightarrow O_2dư\\ n_{O_2\left(dư\right)}=0,04-\dfrac{0,1}{4}=0,015\left(mol\right)\\ \Rightarrow m_{O_2\left(dư\right)}=0,015.32=0,48\left(g\right)\\ b,C1:n_{Na_2O}=\dfrac{2}{4}.n_{Na}=\dfrac{2}{4}.0,1=0,05\left(mol\right)\\ \Rightarrow m_{Na_2O}=0,05.62=3,1\left(g\right)\\ C2:ĐLBTKL:m_{Na_2O}=m_{Na}+m_{O_2\left(bđ\right)}-m_{O_2\left(dư\right)}=2,3+0,04.32-0,48=3,1\left(g\right)\)

a. \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)

\(n_{O_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

PTHH : 4Na + O₂ -t^o> 2Na₂O

             0,1    0,025      0,05

Xét tỉ lệ : \(\dfrac{0,1}{4}< \dfrac{0,04}{1}\) => Na đủ , O₂ dư

\(m_{O_2\left(dư\right)}=\left(0,04-0,025\right).32=0,48\left(g\right)\)

b. Cách 1 : \(m_{Na_2O}=0,05.62=3,1\left(g\right)\)

Cách 2 : \(m_{Na}=0,1.23=2,3\left(g\right)\)

\(m_{O_2}=0,025.32=0,8\left(g\right)\)

Theo ĐLBTKL: 

\(m_{Na}+m_{O_2}=m_{Na_2O}\\ \Rightarrow2,3+0,8=3,1\left(g\right)\)

a,pthh: \(4Na+O_2\underrightarrow{t^o}2Na_2O\)

\(nNa=\dfrac{2,3}{23}=0,1\left(mol\right)\)

\(nO_2=0,896:22,4=0,04\left(mol\right)\)

Xét tỉ lệ : \(\dfrac{nNa}{4}< \dfrac{nO_2}{1}\) ( \(\dfrac{0,1}{4}< \dfrac{0,04}{1}\)) => Oxi dư

\(\Rightarrow nO_{2\left(dư\right)}=0,04-0,025=0,015\left(mol\right)\)

\(nO_{2\left(dư\right)}=0,015.32=0,48\left(g\right)\)

b, do oxi dư nên lấy số mol của natri làm chuẩn.

 \(nNa_2O=0,05\left(mol\right)\)

\(mNa_2O=0,05.\left(23.2+16\right)=3,1\left(gam\right)\)

c2: \(mO_{2\left(đủ\right)}=\) 0,8(gam) 

BTKL : mNa+mO2=mNa2O

<=> 2,3+ 0,8= mNa2O

=>mNa2O= 3,1 (gam)

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

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

a) \(n_{Al}=\dfrac{12,15}{27}=0,45\left(mol\right)\)

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

Xét tỉ lệ: \(\dfrac{0,45}{4}>\dfrac{0,3}{3}\)=> Al dư, O₂ hết

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

            0,4<--0,3-------->0,2

=> \(m_{Al\left(dư\right)}=\left(0,45-0,4\right).27=1,35\left(g\right)\)

b) \(m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)

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

nAl=\(\dfrac{5,4}{27}\)=0,2mol

nO2=\(\dfrac{4,48}{22,4}\)=0,2mol

PTHH:

4Al + 3O2--to->2Al2O3

Tỉ lệ \(\dfrac{0,2}{4}\) <\(\dfrac{0,2}{3}\)->Al hết O2 dưtính theo Al

=>m O2=\(\dfrac{1}{60}\).32=\(\dfrac{8}{15}\)g

2KMnO4-to>K2MnO4+MnO2+O2

0,4--------------------------------------0,2

m KMnO4=0,4.158=63,2g

.

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

Ta có: \(n_{H_2}=n_{O_2}=\dfrac{6,1975}{24,79}=0,25\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,25}{2}< \dfrac{0,25}{1}\), ta được O₂ dư.

Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=0,125\left(mol\right)\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,25-0,125=0,125\left(mol\right)\)

\(\Rightarrow V_{O_2\left(dư\right)}=0,125.24,79=3,09875\left(l\right)\)

b, Theo PT: \(n_{H_2O}=n_{H_2}=0,25\left(mol\right)\)

\(\Rightarrow m_{H_2O}=0,25.18=4,5\left(g\right)\)

c, PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{6}\left(mol\right)\)

\(\Rightarrow m_{KClO_3}=\dfrac{1}{6}.122,5\approx20,42\left(g\right)\)

a. \(n_{H_2}=\dfrac{1.12}{22,4}=0,05\left(mol\right)\)

\(n_{O_2}=\dfrac{2.24}{22,4}=0,1\left(mol\right)\)

PTHH : 2H₂ + O₂ ---t^o---> 2H₂O

            0,05   0,025           0,05

b. Ta thấy : 0,05 < 0,1 => H₂ đủ , O₂ dư

\(V_{O_2\left(dư\right)}=\left(0,1-0,025\right).22,4=1,68\left(l\right)\)

c. \(m_{H_2O}=0,05.18=0,9\left(g\right)\)

Ta có 

nH2 = 1,12 / 22,4 = 0,05 ( mol )

nO2 = 0,1 ( mol )

      2H2 + O2 ---to ---> 2H2O

có : 0,05    0,1

pư : 0,05   0,025          0,05

dư : 0        0,075

Xét tỉ lệ : 0,05 / 2 < 0,1 / 1 , ta được O2 dư

=> nO2 dư = 0,1 - 0,025 = 0,075 ( mol )

=> V = 0,075 . 22,4 = 1,68 ( l )

mH2O = 0,05 . 18 = 0,9 ( g )