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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)
\(\rightarrow m_{O_2}=0,075.32=2,4\left(g\right)\)
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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)
\(1,2H_2+O_2\underrightarrow{t}2H_2O\)
\(2Mg+O_2\underrightarrow{t}2MgO\)
\(2Cu+O_2\underrightarrow{t}2CuO\)
\(S+O_2\underrightarrow{t}SO_2\)
\(4Al+3O_2\underrightarrow{t}2Al_2O_3\)
\(C+O_2\underrightarrow{t}CO_2\)
\(4P+5O_2\underrightarrow{t}2P_2O_5\)
\(2,PTHH:C+O_2\underrightarrow{t}CO_2\)
\(a,n_{O_2}=0,2\left(mol\right)\Rightarrow n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CO_2}=8,8\left(g\right)\)
\(b,n_C=0,3\left(mol\right)\Rightarrow n_{CO_2}=0,3\left(mol\right)\Rightarrow m_{CO_2}=13,2\left(g\right)\)
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.
\(a,n_{O_2}\left(dư\right)=0,05\left(mol\right)\Rightarrow m_{O_2}\left(dư\right)=1,6\left(g\right)\)
\(b,n_{P_2O_5}=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=14,2\left(g\right)\)
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, \(PTHH:4P+5O_2\rightarrow2P_2O_5\)
\(n_P=\frac{46,5}{31}=1,5\left(mol\right)\)
\(\Rightarrow n_{O2}=1,875\left(mol\right)\)
\(\Rightarrow V_{O2}=1,875.22,4=42\left(l\right)\)
\(\Rightarrow V_{kk}=42.5=210\left(l\right)\)
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,\(PTHH:2H_2+O_2\rightarrow2H_2O\)
\(n_{H2}=\frac{33,6}{22,4}=1,5\left(mol\right)\)
\(\Rightarrow n_{O2}=0,75\left(mol\right)\)
\(\Rightarrow V_{O2}=0,75.22,4=16,8\left(l\right)\)
\(\Rightarrow V_{kk}=16,8.5=84\left(l\right)\)
$1)PTHH:C_2H_8O_2+5O_2\xrightarrow{t^o}4CO_2\uparrow+4H_2O$
$n_{C_4H_8O_2}=\dfrac{4,4}{88}=0,05(mol)$
Theo PT: $n_{O_2}=5.0,05=0,25(mol)$
$\Rightarrow V_{O_2}=0,25.22,4=5,6(l)$
$2)PTHH:4Al+3O_2\xrightarrow{t^o}2Al_2O_3$
$n_{O_2}=\dfrac{33,6}{22,4}=1,5(mol)$
Theo PT: $n_{Al}=\dfrac{4}{3}n_{O_2}=2(mol)$
$\Rightarrow m_{Al}=2.27=54(g)$
\(n_{H_2}=\dfrac{V}{24,79}=\dfrac{11,2}{24,79}\approx0,45\left(mol\right)\)
a) \(PTHH:2H_2+O_2\underrightarrow{t^o}2H_2O\)
2 1 2
0,45 0,225 0,45
b) \(m_{O_2}=n.M=0,225.\left(16.2\right)=7,2\left(g\right)\\ V_{O_2}=n.24,79=0,225.24,79=5,57775\left(l\right)\)
c) \(PTHH:2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
2 1 1 1
0,45 0,225 0,225 0,225
\(m_{KMnO_4}=n.M=0,45.\left(39+55+16.4\right)=71,1\left(g\right).\)
a, \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
b, Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=0,25\left(mol\right)\)
\(\Rightarrow m_{O_2}=0,25.32=8\left(g\right)\)
\(V_{O_2}=0,25.22,4=5,6\left(l\right)\)
c, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,5\left(mol\right)\Rightarrow m_{KMnO_4}=0,5.158=79\left(g\right)\)
Câu 1 :
Phản ứng phân hủy : Xem SGK .
Giải thích : Khi tắt đèn cồn người ta đậy nắp đèn lại là vì để không cung cấp tiếp khí oxi cho đèn. Khi oxi hết , đèn sẽ tự tắt.
Câu 2 :
Bốn công thức hóa học của oxit axit:
SO2: Lưu huỳnh đioxit.
P2O5: điphotpho pentaoxit
N2O5: đinito pentaoxit.
CO2: cacbon dioxit.
- Bốn oxit bazo:
K2O: kali oxit
Na2O: natri oxit
CaO: canxi oxit
Al2O3: nhôm oxit
Câu 3 :
a, \(n_P=\frac{46,5}{31}=1,5\left(mol\right)\)
\(4P+5O_2\underrightarrow{^{to}}2P_2O_5\)
\(\Rightarrow n_{O2}=\frac{5}{4}n_P=\frac{5}{4}.1,5=1,875\left(mol\right)\)
\(\Rightarrow m_{O2}=1,875.32=60\left(g\right)\)
b,\(n_{Al}=\frac{67,5}{27}=2,5\left(mol\right)\)
\(4Al+3O_2\underrightarrow{^{to}}2Al_2O_3\)
\(\Rightarrow n_{O2}=\frac{3}{4}n_{Al}=\frac{3}{4}.2,5=1,875\left(mol\right)\)
\(\Rightarrow m_{O2}=1,875.32=60\left(g\right)\)
Câu 4 :
a, \(n_{SO2}=\frac{19,2}{32+16.2}=0,3\left(mol\right)\)
\(n_{O2}=0,46875\left(mol\right)\)
Nên O2 dư , S hết
\(\Rightarrow n_S=n_{SO2}=0,3\left(mol\right)\)
\(\Rightarrow m_S=9,6\left(g\right)\)
b,\(n_{O2\left(pư\right)}=n_{SO2}=0,3\left(mol\right)\)
\(\Rightarrow n_{O2\left(dư\right)}=0,46875-0,3=0,16875\left(mol\right)\)
\(\Rightarrow m_{O2\left(dư\right)}=5,4\left(g\right)\)
Bài này bạn viết phương trình rồi tính số mol các chất cho sẵn => Tính số mol chất yêu cầu ( oxi ) => Rồi tính khối lượng oxi cần dùng thôi bạn...