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Tính khối lượng khí oxi cần dùng để đốt cháy hết hỗn hợp:
Mình cho a,b,c,d,... nha!
a) 0,5 mol Fe.
PTHH: 3Fe + 2O2 -to-> Fe3O4
Theo PTHH và đề bài, ta có:
\(n_{O_2}=\frac{2.n_{Fe}}{3}=\frac{2.0,5}{3}\approx0,333\left(mol\right)\)
=> \(m_{O_2}=0,333.32=10,656\left(g\right)\)
b) 1,25 mol nhôm
PTHH: 4Al + 3O2 -to-> 2Al2O3
Theo PTHH và đề bài, ta có:
\(n_{O_2}=\frac{3.1,25}{4}=0,9375\left(mol\right)\)
=> \(m_{O_2}=32.0,9375=30\left(g\right)\)
c) 1,5 mol Zn
2Zn + O2 -to-> 2ZnO
Theo PTHH và đề bài, ta có:
\(n_{O_2}=\frac{1,5}{2}=0,75\left(mol\right)\)
=> \(m_{O_2}=0,75.32=24\left(g\right)\)
d) Ta có:
\(n_P=\frac{3,1}{31}=0,1\left(mol\right)\)
PTHH: 4P + 5O2 -to-> 2P2O5
Theo PTHH và đề bài, ta có:
\(n_{O_2}=\frac{5.0,1}{4}=0,125\left(mol\right)\)
=> \(m_{O_2}=0,125.32=4\left(g\right)\)
e) Ta có:
\(n_S=\frac{6,4}{32}=0,2\left(mol\right)\)
PTHH: S + O2 -to-> SO2
Theo PTHH và đề bài, ta có:
\(n_{O_2}=n_S=0,2\left(mol\right)\\ =>m_{O_2}=0,2.32=6,4\left(g\right)\)
f) Ta có:
\(n_C=\frac{3,6}{12}=0,3\left(mol\right)\)
PTHH: C + O2 -to-> CO2
Theo PTHH và đề bài, ta có:
\(n_{O_2}=n_C=0,3\left(mol\right)\)
=> \(m_{O_2}=32.0,3=9,6\left(g\right)\)
Với 3 mol hỗn hợp khí có : 1.5 (mol) H2 , 0.5 (mol) N2 và 1 (mol) CO2
\(\overline{M}=\dfrac{1.5\cdot2+0.5\cdot28+1\cdot44}{3}=20.33\left(g\text{/}mol\right)\)
a) PTHH :
2KClO3 \(\rightarrow\)2KCl + 3O2 (1)
CH4 + 2O2 \(\rightarrow\)CO2 + 2H2O (2)
2H2 + O2 \(\rightarrow\) 2H2O (3)
Theo PT(2) => nO2 = 2.nCH4 = 2 x 0.5 =1(mol)
Theo PT(3) => nO2 = 1/2 x nH2 = 1/2 x 0.25 =0.125(mol)
=> tổng nO2 = 1+ 0.125 =1.125(mol)
Theo PT(1) => nKClO3 = 2/3 . nO2 = 2/3 x 1.125 = 0.75(mol)
=> mKClO3 = n .M = 0.75 x 122.5 =91.875(g)
b) 4Al + 3O2 \(\rightarrow\) 2Al2O3 (4)
2Zn + O2 \(\rightarrow\) 2ZnO (5)
nAl = m : M = 6.75/27=0.25(mol)
nZn = m/M = 9.75/65 =0.15(mol)
Theo PT(4) => nO2 = 3/4 . nAl = 3/4 x 0.25 =0.1875(mol)
Theo PT(5) => nO2 = 1/2 x nZn = 1/2 x 0.15 =0.075(mol)
tổng nO2 = 0.1875 + 0.075 =0.2625(mol)
theo PT(1) => nKClO3 = 2/3 x nO2 = 2/3 x 0.2625 =0.175(mol)
=> mKClO3 = n .M = 0.175 x 122.5 =21.4375(g)
a) \(CH_4+2O_2\underrightarrow{t\text{°}}CO_2+2H_2O\)(1)
___0,5------>1___________________(mol)
\(2H_2+O_2\underrightarrow{t\text{°}}2H_2O\)(2)
0,25->0,125__0,25____(mol)
\(2KClO_3\underrightarrow{t\text{°}}2KCl+3O_2\)(3)
0,75<---------------0,125+1=1,125(mol)
=> mKClO3= 0,75*122,5=91,875(g)
b) n Al= 6,75/27=0,25(mol)
n Zn= 9,75/65= 0,15 (mol)
\(4Al+2O_2\underrightarrow{t\text{°}}2Al_2O_3\) (1')
0.25-->0.125_________(mol)
\(2Zn+O_2\underrightarrow{t\text{}\text{°}}2ZnO\) (2')
0.15->0.075_________(mol)
\(2KClO_3\underrightarrow{t\text{°}}2KCl+3O_2\)(3')
\(\frac{2}{15}\)<-------------------0.075+0.125=0.2(mol)
m KClO3=\(\frac{2}{15}\)*122,5=\(\frac{49}{3}\) (g)
\(a,Đặt:n_{CH_4}=a\left(mol\right);n_{C_4H_{10}}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ 2C_4H_{10}+13O_2\rightarrow\left(t^o\right)8CO_2+10H_2O\\ \Rightarrow\left\{{}\begin{matrix}16a+58b=7,4\\22,4a+22,4.4b=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{CH_4}=0,1.16=1,6\left(g\right)\\m_{C_4H_{10}}=0,1.58=5,8\left(g\right)\end{matrix}\right.\\ b,n_{O_2}=2a+\dfrac{13}{2}b=2.0,1+6,5.0,1=0,85\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,85.22,4=19,04\left(l\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\x_{Ca}=y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}24x+40y=17,6\\x=2y\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,4\\y=0,2\end{matrix}\right.\)
a)\(m_{Mg}=0,4\cdot24=9,6g\)
\(m_{Ca}=0,2\cdot40=8g\)
b)\(2Mg+O_2\underrightarrow{t^o}2MgO\)
\(2Ca+O_2\underrightarrow{t^o}2CaO\)
Từ hai pt: \(\Rightarrow\Sigma n_{O_2}=\dfrac{1}{2}n_{Mg}+\dfrac{1}{2}n_{Ca}=\dfrac{1}{2}\cdot0,4+\dfrac{1}{2}\cdot0,2=0,3mol\)
\(\Rightarrow m_{O_2}=0,3\cdot32=9,6g\)
\(V_{O_2}=0,3\cdot22,4=6,72l\)
\(\Rightarrow V_{kk}=5V_{O_2}=5\cdot6,72=33,6l\)
a)
Có \(\left\{{}\begin{matrix}24.n_{Mg}+40.n_{Ca}=17,6\\n_{Mg}=2.n_{Ca}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{Ca}=0,2\left(mol\right)\\n_{Mg}=0,4\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}m_{Ca}=0,2.40=8\left(g\right)\\m_{Mg}=0,4.24=9,6\left(g\right)\end{matrix}\right.\)
b)
PTHH: 2Ca + O2 --to--> 2CaO
0,2-->0,1
2Mg + O2 --to--> 2MgO
0,4--->0,2
=> \(V_{O_2}=\left(0,1+0,2\right).22,4=6,72\left(l\right)\)
\(V_{kk}=6,72.5=33,6\left(l\right)\)
b, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
\(n_P=\dfrac{6,2}{31}=0,2mol\\n_{O_2}=\dfrac{0,2.5}{4}=0,25mol \)
\(S+O_2\underrightarrow{t^o}SO_2\)
\(n_S=\dfrac{3,2}{32}=0,1mol\\ n_{O_2}=0,1mol\)
\(C+O_2\underrightarrow{t^o}CO_2\)
\(n_C=\dfrac{2,4}{12}=0,2mol\\ n_{O_2}=0,2mol\\ n_{O_2}\left(tổng\right)=\)
\(0,25+0,1+0,2=0,55mol\\ m_{O_2}\left(trong.hh.B\right)=0,55.32=17,6g\)
a, \(m_{Fe}=0,25.56=14g\)
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
\(\Rightarrow n_{O_2}=\dfrac{0,25.2}{3}=0,16mol\\ m_{O_2}=0,16.32=5,12g\)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
\(n_{O_2}=\dfrac{0,25.3}{4}=0,1875mol\\ m_{O_2}=0,1875.32=6g\)
\(2Zn+O_2\underrightarrow{t^o}2ZnO\)
\(n_{O_2}=\dfrac{0,5.1}{2}=0,25mol\\ m_{O_2}=0,25.32=8g\)
\(\Rightarrow m_{O_2}\left(trong.hỗn.hợp.A\right)=\) \(5,12+6+8=19,12g\)