nO2=48/32=1,5(mol)

a) PTHH: C+ O2 -to-> CO2

nC=nO2=1,5(mol)

=> mC=nC.M(C)=1,5.12=18(g)

b) PTHH: S+ O2 -to-> SO2

nS=nO2=1,5(mol)

=> mS=nS.M(S)=1,5.32= 48(g)

c) PTHH: 4P + 5 O2 -to->2  P2O5

nP= 4/5. nO2=4/5 . 1,5=1,2(mol)

=>mP=1,2.31=37,2(g)

Chúc em học tốt!

\(n_{O_2}=\dfrac{48}{32}=1.5\left(mol\right)\)

\(a.\)

\(C+O_2\underrightarrow{^{^{t^0}}}CO_2\)

\(n_C=n_{O_2}=1.5\left(mol\right)\)

\(m_C=1.5\cdot12=18\left(g\right)\)

\(b.\)

\(S+O_2\underrightarrow{^{^{t^0}}}SO_2\)

\(n_S=n_{O_2}=1.5\left(mol\right)\)

\(m_S=1.5\cdot32=48\left(g\right)\)

\(c.\)

\(4P+5O_2\underrightarrow{^{^{t^0}}}2P_2O_5\)

\(n_P=\dfrac{4}{5}\cdot n_{O_2}=\dfrac{4}{5}\cdot1.5=1.2\left(mol\right)\)

\(m_P=1.2\cdot31=37.2\left(g\right)\)

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

a) PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

Theo PTHH: \(n_{Fe}=\dfrac{3}{2}n_{O_2}=0,06\left(mol\right)\) \(\Rightarrow m_{Fe}=0,06\cdot56=3,36\left(g\right)\)

b và c tương tự

d) PTHH: \(4FeS+7O_2\underrightarrow{t^o}2Fe_2O_3+4SO_2\)

The PTHH: \(n_{FeS}=\dfrac{4}{7}n_{O_2}=\dfrac{4}{175}\left(mol\right)\)

\(\Rightarrow m_{FeS}=\dfrac{4}{175}\cdot88\approx2,01\left(g\right)\)

\(1,2H_2+O_2\underrightarrow{t}2H_2O\)

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

\(2Cu+O_2\underrightarrow{t}2CuO\)

\(S+O_2\underrightarrow{t}SO_2\)

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

\(C+O_2\underrightarrow{t}CO_2\)

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

\(2,PTHH:C+O_2\underrightarrow{t}CO_2\)

\(a,n_{O_2}=0,2\left(mol\right)\Rightarrow n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CO_2}=8,8\left(g\right)\)

\(b,n_C=0,3\left(mol\right)\Rightarrow n_{CO_2}=0,3\left(mol\right)\Rightarrow m_{CO_2}=13,2\left(g\right)\)

c, Vì\(\frac{0,3}{1}>\frac{0,2}{1}\)nên C phản ửng dư, O₂ phản ứng hết, Bài toán tính theo O₂

\(n_{O_2}=0,2\left(mol\right)\Rightarrow n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CO_2}=8,8\left(g\right)\)

\(3,PTHH:CH_4+2O_2\underrightarrow{t}CO_2+2H_2O\)

\(C_2H_2+\frac{5}{2}O_2\underrightarrow{t}2CO_2+H_2O\)

\(C_2H_6O+3O_2\underrightarrow{t}2CO_2+3H_2O\)

\(4,a,PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)

\(n_P=1,5\left(mol\right)\Rightarrow n_{O_2}=1,2\left(mol\right)\Rightarrow m_{O_2}=38,4\left(g\right)\)

\(b,PTHH:C+O_2\underrightarrow{t}CO_2\)

\(n_C=2,5\left(mol\right)\Rightarrow n_{O_2}=2,5\left(mol\right)\Rightarrow m_{O_2}=80\left(g\right)\)

\(c,PTHH:4Al+3O_2\underrightarrow{t}2Al_2O_3\)

\(n_{Al}=2,5\left(mol\right)\Rightarrow n_{O_2}=1,875\left(mol\right)\Rightarrow m_{O_2}=60\left(g\right)\)

\(d,PTHH:2H_2+O_2\underrightarrow{t}2H_2O\)

\(TH_1:\left(đktc\right)n_{H_2}=1,5\left(mol\right)\Rightarrow n_{O_2}=0,75\left(mol\right)\Rightarrow m_{O_2}=24\left(g\right)\)

\(TH_2:\left(đkt\right)n_{H_2}=1,4\left(mol\right)\Rightarrow n_{O_2}=0,7\left(mol\right)\Rightarrow m_{O_2}=22,4\left(g\right)\)

\(5,PTHH:S+O_2\underrightarrow{t}SO_2\)

\(n_{O_2}=0,46875\left(mol\right)\)

\(n_{SO_2}=0,3\left(mol\right)\)

Vì\(0,46875>0,3\left(n_{O_2}>n_{SO_2}\right)\)nên S phản ứng hết, bài toán tính theo S.

\(a,\Rightarrow n_S=n_{SO_2}=0,3\left(mol\right)\Rightarrow m_S=9,6\left(g\right)\)

\(n_{O_2}\left(dư\right)=0,16875\left(mol\right)\Rightarrow m_{O_2}\left(dư\right)=5,4\left(g\right)\)

\(6,a,PTHH:C+O_2\underrightarrow{t}CO_2\)

\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_C=1,5\left(mol\right)\Rightarrow m_C=18\left(g\right)\)

\(b,PTHH:2H_2+O_2\underrightarrow{t}2H_2O\)

\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_{H_2}=0,75\left(mol\right)\Rightarrow m_{H_2}=1,5\left(g\right)\)

\(c,PTHH:S+O_2\underrightarrow{t}SO_2\)

\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_S=1,5\left(mol\right)\Rightarrow m_S=48\left(g\right)\)

\(d,PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)

\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_P=1,2\left(mol\right)\Rightarrow m_P=37,2\left(g\right)\)

\(7,n_{O_2}=5\left(mol\right)\Rightarrow V_{O_2}=112\left(l\right)\left(đktc\right)\);\(V_{O_2}=120\left(l\right)\left(đkt\right)\)

\(8,PTHH:C+O_2\underrightarrow{t}CO_2\)

\(m_C=0,96\left(kg\right)\Rightarrow n_C=0,08\left(kmol\right)=80\left(mol\right)\Rightarrow n_{O_2}=80\left(mol\right)\Rightarrow V_{O_2}=1792\left(l\right)\)

\(9,n_p=0,2\left(mol\right);n_{O_2}=0,3\left(mol\right)\)

\(PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)

Vì\(\frac{0,2}{4}< \frac{0,3}{5}\)nên P hết O2 dư, bài toán tính theo P.

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

\(b,n_{P_2O_5}=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=14,2\left(g\right)\)

đủ cả 9 câu bạn nhé,

a) S + O₂ --t^o--> SO₂

b) \(n_S=\dfrac{1,6}{32}=0,05\left(mol\right)\)

PTHH: S + O₂ --t^o--> SO₂

        0,05-->0,05--->0,05

=> V_SO2 = 0,05.22,4 = 1,12 (l)

c) 

PTHH: 2KClO₃ --t^o--> 2KCl + 3O₂

             \(\dfrac{1}{30}\)<-----------------0,05

=> \(m_{KClO_3}=\dfrac{1}{30}.122,5=\dfrac{49}{12}\left(g\right)\)

n O2=33,6/22,4=1,5(mol)

a) C+O2------>CO2

1,5<--1,5

m C=1,5.12=18(g)

b) 2H2+O2----->2H2O

3<------1,5

m H2=3.2=6(g)

c) S+O2----->SO2

1,5<----1,5

m S=1,5.32=48(g)

d) 4P+5O2--->2P2O5

1,2<----1,5

m P=1,2.31=37,2(g)

+n_O2 = 33,6 /22,4 = 1,5 (mol)

A. C + O₂ -----------> CO₂ (1)

Theo (1) : n_C = n_O2 = 1,5 (mol)

=> m _C = 1,5 .12 = 18 (g)

B. 2H₂ + O₂ --------> 2H₂O (2)

Theo (2) : n_H2 = 2n_O2 = 1,5.2=3 (mol)

=> m _H2 = 3.2 = 6 g

C. S + O₂ -------> SO₂ (3)

Theo (3) : n_S = n_O2 = 1,5 (mol)

=> m_S= 1,5 . 32 = 48 (g)

D. 4P +5O₂ -----------> 2P₂O₅ (4)

Theo (4) : n_P = nO₂ . 4/5 = 1,5 .4/5 = 1,2 (mol)

=> m_P = 31 .1,2 = 37,2 (g)

câu 1 một bình chứa 33,6 lít khí oxi (điều kiện tiêu chuẩn) với thể tích này có thể đốt cháy:

    a) bao nhiêu game cacbon và tạo ra bao nhiêu lít cacbon dioxit

    b) bao nhiêu gam lưu huỳnh và tạo ra bao nhiêu lít lưu huỳnh dioxit

    c) bao nhiêu gam P và tạo ra bao nhiêu gam diphotpho pentaoxit

\(n_{KMnO_4}=\dfrac{15.8}{158}=0.1\left(mol\right)\)

\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)

\(n_{O_2}=\dfrac{0.1}{2}=0.05\left(mol\right)\)

\(V_{O_2}=0.05\cdot22.4=1.12\left(l\right)\)

\(b.\)

\(n_P=\dfrac{6.2}{31}=0.2\left(mol\right)\)

\(4P+5O_2\underrightarrow{t^0}2P_2O_5\)

\(4.........5\)

\(0.2........0.05\)

\(LTL:\dfrac{0.2}{4}>\dfrac{0.05}{5}\Rightarrow Pdư\)

\(m_{P\left(dư\right)}=\left(0.2-0.04\right)\cdot31=4.96\left(g\right)\)

PTHH:2KMnO4--- K2MnO4+MnO2 +O2

ADCT nKmno4=15,8/158=0,1 mol 

a, theo pt có nO2/nKmno4= 1/2 

nO2=0,05 mol 

ADCT V=n*22,4 

VO2=0,05*22,4 =1,12 l

b, PTHH: 5O2+4P---2P2O5

ADCTnP=6,2/31=0,2 mol 

Theo pt 

nO2/5=0,01     bé hơn     nP/4=0,05 

P dư 

theo pt nP(pư)/nO2=4/5 

nP(p/ư)=0,04 mol 

nP(dư)=0,05-0,04 =0,01 mol

ADCT:m=n*M

mP(dư)=0,01*31=0,31g

nO2= 1,5(mol)

PTHH: 4P + 5 O2 -to-> 2 P2O5

nP= 4/5.nO2= 4/5.1,5= 1,2(mol)

=>mP= 1,2. 31= 37,2(g)

---

S+ O2-to-> SO2

nS= nO2= 1,5(mol)

=>mS= 1,5.32= 48(g)

---

C+ O2-to->CO2

nC= nO2= 1,5(mol)

=>mC=1,5.12=18(G)

---

3 Fe+ 2 O2 -to-> Fe3O4

nFe=3/2. nO2= 3/2 .1,5=2,25(mol)

=>mFe= 2,25.56=126(g)

---

2 Zn+ O2 -to->ZnO

=> nZn=2.nO2= 2.1,5=3(mol)

=>mZn=3.65=195(g)

\(n_{O2}=\frac{33,6}{22,4}=1,5\left(mol\right)\)

\(4P+5O_2\rightarrow2P_2O_5\)

\(n_P=\frac{5}{4}.n_{O2}=\frac{5}{4}.1,5=1,875\left(mol\right)\)

\(S+O_2\rightarrow SO_2\)

\(n_S=n_{O2}=1,5\left(mol\right)\)

\(C+O_2\rightarrow CO_2\)

\(n_C=n_{O2}=1,5\left(mol\right)\)

\(3Fe+2O_2\rightarrow Fe_3O_4\)

\(n_{Fe}=\frac{3}{2}.n_{O2}=2,25\left(mol\right)\)

\(\Rightarrow m_{Fe}=2,25.56=126\left(g\right)\)

\(2Zn+O_2\rightarrow2ZnO\)

\(n_{Zn}=n_{O2}=1,5\left(mol\right)\)

\(\Rightarrow m_{Zn}=1,5.65=97,5\left(g\right)\)