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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...
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=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 :))
$a)4P+5O_2\xrightarrow{t^o}2P_2O_5$
$n_P=\dfrac{4,65}{31}=0,15(mol)$
$\Rightarrow n_{O_2}=0,1875(mol)$
$\Rightarrow m_{O_2}=0,1875.32=6(g)$
$b)4Al+3O_2\xrightarrow{t^o}2Al_2O_3$
$n_{Al}=\dfrac{6,75}{27}=0,25(mol)$
$\Rightarrow n_{O_2}=0,1875(mol)$
$\Rightarrow m_{O_2}=0,1875.32=6(g)$
$c)2H_2+O_2\xrightarrow{t^o}2H_2O$
$n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)$
$\Rightarrow n_{O_2}=0,075(mol)$
$\Rightarrow m_{O_2}=0,075.32=2,4(g)$
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,\(PTHH:2H_2+O_2\rightarrow2H_2O\)
\(n_{H2}=\frac{33,6}{22,4}=1,5\left(mol\right)\)
\(\Rightarrow n_{O2}=0,75\left(mol\right)\)
\(\Rightarrow V_{O2}=0,75.22,4=16,8\left(l\right)\)
\(\Rightarrow V_{kk}=16,8.5=84\left(l\right)\)
\(1,2H_2+O_2\underrightarrow{t}2H_2O\)
\(2Mg+O_2\underrightarrow{t}2MgO\)
\(2Cu+O_2\underrightarrow{t}2CuO\)
\(S+O_2\underrightarrow{t}SO_2\)
\(4Al+3O_2\underrightarrow{t}2Al_2O_3\)
\(C+O_2\underrightarrow{t}CO_2\)
\(4P+5O_2\underrightarrow{t}2P_2O_5\)
\(2,PTHH:C+O_2\underrightarrow{t}CO_2\)
\(a,n_{O_2}=0,2\left(mol\right)\Rightarrow n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CO_2}=8,8\left(g\right)\)
\(b,n_C=0,3\left(mol\right)\Rightarrow n_{CO_2}=0,3\left(mol\right)\Rightarrow m_{CO_2}=13,2\left(g\right)\)
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.
\(a,n_{O_2}\left(dư\right)=0,05\left(mol\right)\Rightarrow m_{O_2}\left(dư\right)=1,6\left(g\right)\)
\(b,n_{P_2O_5}=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=14,2\left(g\right)\)
\(a.\)
\(n_P=\dfrac{46.5}{31}=1.5\left(mol\right)\)
\(4P+5O_2\underrightarrow{t^0}2P_2O_5\)
\(1.5...1.875\)
\(m_{O_2}=1.875\cdot32=60\left(g\right)\)
\(b.\)
\(n_{Al}=\dfrac{67.4}{27}\simeq2.5\left(mol\right)\)
\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)
\(2.5.......1.875\)
\(m_{O_2}=1.875\cdot32=60\left(g\right)\)
\(c.\)
\(n_{H_2}=\dfrac{33.6}{22.4}=1.5\left(mol\right)\)
\(2H_2+O_2\underrightarrow{t^0}2H_2O\)
\(1.5....0.75\)
\(m_{O_2}=0.75\cdot32=24\left(g\right)\)
Chúc em học tốt !!
a. PT: 4P + 5O2 -----> 2P2O5.
Ta có: nP=46,5/31=1,5(mol).
Theo PT, ta có: nO2= 5/4 . 1,5=1,875(mol).
=> mO2= 1,875.32=60(g).
b.PT: 4Al + 3O2 -----> 2Al2O3.
Ta có: nAl= 67,4/27=2,5(mol).
Theo PT, ta có: nO2= 3/4 . 2,5 =1,875(mol)
=> mO2= 1,875.32=60(g)
c. PT: 2H2 + O2 -----> 2H2O.
Ta có: nH2= 33,6/22,4=1,5(mol)
Theo PT, ta có: nO2= 1/2 . 1,5 =0,75 (mol).
=> mO2= 0,75.32=24(g)
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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)