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

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

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

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

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

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

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

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

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

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

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,

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

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

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

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

1,5________1,5(mol)

\(m_C=1,5.12=18\left(g\right)\)

b) \(2H_2+O_2\underrightarrow{t^O}2H_2O\)

3________1,5_____3(mol)

\(m_{H2}=3.2=6\left(g\right)\)

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

1,5__________1,5(mol)

\(m_S=1,5.32=48\left(g\right)\)

Chúc bạn học tốt

\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\\ n_{O_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ pthh:S+O_2\underrightarrow{t^o}SO_2\\ LTL:\dfrac{0,1}{1}>\dfrac{0,05}{1}\) 
=> S dư 
\(n_{S\left(P\text{Ư}\right)}=n_{SO_2}=n_{O_2}=0,05\left(mol\right)\\ m_S=\left(0,1-0,05\right).32=1,6\left(g\right)\\ V_{SO_2}=0,05.22,4=1,12\left(l\right)\)

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

\(a) n_P = \dfrac{18,6}{31} = 0,6(mol)\\ n_{O_2} = \dfrac{20,16}{22,4} = 0,9(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ \dfrac{n_P}{4} = 0,15 < \dfrac{n_{O_2}}{5} = 0,18 \to O_2\ dư\\ n_{O_2\ pư} = \dfrac{5}{4}n_P = 0,75(mol)\\ \Rightarrow m_{O_2\ dư} = (0,9-0,75).32 = 4,8(gam)\\ b) n_{Fe} = \dfrac{56}{56} = 1(mol)\)

\(3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ \dfrac{n_{Fe}}{3} = \dfrac{1}{3}<\dfrac{n_{O_2}}{2} = 0,45\to Fe\ cháy\ hết.\\ c)\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,9.2 = 1,8(mol)\\ \Rightarrow m_{KMnO_4} = 1,8.158 =284,4(gam)\)

Mọi người giải luôn hộ nhé

ta tính nO2=44,8:22,4=2mol

          4<-2

=> cần 4 mol Cacbon

                    8/5<-2

vậy cần 1,6mol Photpho

b) Fe+O2=>FeO

    2<-2

=> cần 2 mol bột sắt