a) \(n_P=\dfrac{46,5}{31}=1,5\left(mol\right)\)

\(4P+5O_2\rightarrow2P_2O_5\)

1,5.....1,875 (mol) 

\(\rightarrow m_{O_2}=1,875.32=60\left(g\right)\)

c. \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(2H_2+O_2\rightarrow2H_2O\)

0,15.....0,075 (mol) 

\(\rightarrow m_{O_2}=0,075.32=2,4\left(g\right)\)

a, \(n_P=\dfrac{m}{M}=1,5\left(mol\right)\)

\(PTHH:4P+5O_2\rightarrow2P_2O_5\)

...............1,5......1,875......

=> \(m_{O_2}=n.M=60\left(g\right)\)

b, \(n_{Al}=\dfrac{m}{M}=2,5\left(mol\right)\)

\(PTHH:4Al+3O_2\rightarrow2Al_2O_3\)

................2,5.....1,875.......

=> \(m_{O_2}=n.M=60\left(g\right)\)

c, ( Chắc là ở ĐKTC )

\(n_{H_2}=\dfrac{V}{22,4}=1,5\left(mol\right)\)

\(PTHH:2H_2+O_2\rightarrow2H_2O\)

...............1,5....0,75...........

=> \(m_{O_2}=n.M=24\left(g\right)\)

Chép có mà mỏi tay:https://hoc24.vn/hoi-dap/tim-kiem?q=t%C3%ADnh+l%C6%B0%E1%BB%A3ng+kh%C3%AD+oxi+c%E1%BA%A7n+d%C3%B9ng+%C4%91%E1%BB%83+%C4%91%E1%BB%91t+ch%C3%A1y+h%E1%BA%BFt++a/+46,5g+photpho++b/+67,5g+nh%C3%B4m++c/+30g+cacbon++d/+33,6+l+kh%C3%AD+hidro&id=177388

a) nP= 46,5/31= 1,5(mol)

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

1,5________1,875(mol)

=>mO2= 1,875.32= 60(g)

b) nAl= 67,5/27= 2,5(mol)

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

2,5______1,875(mol)

=> mO2= 1,875.32= 60(g)

c) nH2 = 33,6/22,4= 1,5(mol)

PTHH: H2 + 1/2 O2 -to-> H2O

1,5________0,75(mol)

=> mO2= 0,75.32= 24(g)

a) Pthh: \(P+O_2\rightarrow P_2O_5\)

\(\Rightarrow4P+5O_2\rightarrow2P_2O_5\)

số mol của 46,5g P là:

\(n=\frac{m}{M}=\frac{31}{46,5}=0,6\)

Số mol của oxi là:\(\frac{0,6.5}{4}=0,75\left(mol\right)\)

KL oxi cần dùng là

\(m=n.M=0,75.32=24\left(g\right)\)

PTHH: C₂H₄ (0,5 mol) + 3O₂ (1,5 mol) \(\underrightarrow{t^o}\) 2CO₂ (1 mol) + 2H₂O.

Thể tích khí oxi cần dùng ở đktc: V_{khí oxi}=1,5.22,4=33,6 (lít).

Khối lượng khí CO₂ sinh ra là: m_{khí cacbonic}=1.44=44 (g).

Chọn A.

A

\(n_{C_2H_4}=\dfrac{14}{28}=0,5mol\)

\(C_2H_4+5O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)

 0,5        2,5                1                      ( mol )

\(V_{O_2}=2,5.22,4=56l\)

\(m_{CO_2}=1.44=44g\)

( ko có câu đúng :< )

a) n P=46,5/31=1,5(mol)

4P+5O2-->2P2O5

1,5---1,875

m O2=1,875.32=60(g)

b) n C=30/12=2,5(g)

C+O2--->CO2

2,5--2,5

m O2=2,5.32=80(g)

c) m Al=67,5/27=2,5(mol)

4Al+3O2--->2Al2O3

2,5----1,875(mol)

m O2=0,1875.32=60(g)

d) n H2=33,6/22,4=1,5(mol)

2H2+O2-->2H2O

1,5-----0,75(mol)

m O2=0,75.32=24(g)

Chúc bạn học tốt :))

a)

nP= 46,5/31 = 1,5(mol)

4P + 5O2 --> 2P2O5

m O2=1,875.32=60(g)

b)

nC= 30/12 = 2,5(mol)

C + O2 -->CO2

mO2=2,5.32= 80 (g)

c)

nAl= 67,5/27=2,5 (mol)

4Al + 3O2 --> 2Al2O3

mO2= 0,1875.32 =60 (g)

d) nH2=33,6/22,4=1,5(mol)

2H2 + O2 --> 2H2O

mO2=0,75.32=24(g)

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

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

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

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

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

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

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

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

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

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

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, \(PTHH:4P+5O_2\rightarrow2P_2O_5\)

\(n_P=\frac{46,5}{31}=1,5\left(mol\right)\)

\(\Rightarrow n_{O2}=1,875\left(mol\right)\)

\(\Rightarrow V_{O2}=1,875.22,4=42\left(l\right)\)

\(\Rightarrow V_{kk}=42.5=210\left(l\right)\)

b,\(PTHH:C+O_2\rightarrow CO_2\)

\(n_C=\frac{30}{12}=2,5\left(mol\right)\)

\(\Rightarrow n_{O2}=n_C=2,5\left(mol\right)\)

\(\Rightarrow V_{O2}=2,5.22,4=56\left(l\right)\)

\(\Rightarrow V_{kk}=56.5=280\left(l\right)\)

c,\(PTHH:4Al+3O_2\rightarrow2Al_2O_3\)

\(n_{Al}=\frac{67,5}{27}=2,5\left(mol\right)\)

\(\Rightarrow n_{O2}=1,875\left(mol\right)\)

\(\Rightarrow V_{O2}=1,875.22,4=42\left(l\right)\)

\(\Rightarrow V_{kk}=42.5=210\left(l\right)\)

d,\(PTHH:2H_2+O_2\rightarrow2H_2O\)

\(n_{H2}=\frac{33,6}{22,4}=1,5\left(mol\right)\)

\(\Rightarrow n_{O2}=0,75\left(mol\right)\)

\(\Rightarrow V_{O2}=0,75.22,4=16,8\left(l\right)\)

\(\Rightarrow V_{kk}=16,8.5=84\left(l\right)\)

PTHH : \(C_2H_4\rightarrow2H_2O+2CO_2\)

\(n_{C_2H_4}=\frac{14}{28}=0,5\left(mol\right)\)

Theo phương trình : 

\(n_{O_2}=3n_{C_2H_4}=3.0,5=1,5\left(mol\right)\)

\(\Rightarrow V_{O_2}=1,5.22,4=33,6\left(l\right)\)

\(n_{CO_2}=2n_{C_2H_4}=2.0,5=1\left(mol\right)\)

\(\Rightarrow m_{CO_2}=1.44=44\left(gam\right)\)

PTHH: C2H4 (0,5 mol) + 3O2 (1,5 mol) to→ 2CO2 (1 mol) + 2H2O.

Thể tích khí oxi cần dùng ở đktc: Vkhí oxi=1,5x22,4=33,6 (lít).

Khối lượng khí CO2 sinh ra là: mkhí cacbonic=1x44=44 (gam).

Chọn A.