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\(n_S=\dfrac{3.2}{32}=0.1\left(mol\right)\)
\(n_{O_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(S+O_2\underrightarrow{t^0}SO_2\)
\(0.05.0.05...0.05\)
\(\Rightarrow Sdư\)
\(V_{SO_2}=0.05\cdot22.4=1.12\left(l\right)\)
\(b.\)
\(S+O_2\underrightarrow{t^0}SO_2\)
\(0.1..0.1\)
\(V_{kk}=5V_{O_2}=5\cdot0.1\cdot22.4=11.2\left(l\right)\)
a, PT: \(S+O_2\underrightarrow{t^o}SO_2\)
Ta có: \(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,05}{1}\), ta được S dư.
Theo PT: \(n_{SO_2}=n_{O_2}=0,05\left(mol\right)\) \(\Rightarrow V_{SO_2}=0,05.22,4=1,12\left(l\right)\)
b, Theo PT: \(n_{O_2}=n_S=0,1\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)
\(\Rightarrow V_{kk}=2,24.5=11,2\left(l\right)\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ n_{CH_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{CO_2}=n_{CH_4}=0,2\left(mol\right)\\ n_{O_2}=2.n_{CH_4}=2.0,2=0,4\left(mol\right)\\ a,V_{kk}=5.V_{O_2\left(đktc\right)}=5.\left(0,4.22,4\right)=44,8\left(l\right)\\ b,m_{CO_2}=0,2.44=8,8\left(g\right)\)
\(V_{CH_4(nguyên chất)}=1000.98\%=980(l)\\ \Rightarrow n_{CH_4}=\dfrac{980}{22,4}=43,75(mol)\\ CH_4+2O_2\xrightarrow{t^o}CO_2+2H_2O\\ \Rightarrow n_{O_2}=87,5(mol)\\ \Rightarrow V_{O_2}=87,5.22,4=1960(l)=1,96(m^3)\)
S + O2 →SO2
a) nO2 = 2,24/22,4 = 0,1 mol
=> nSO2 = 0,1 mol
<=> V SO2 = 0,1 .22,4 = 2,24 lít
b) nS = O2 = 0,1 mol
=> mS = 0,1.32 = 3,2 gam
S + O2 →SO2
a) nO2 = 2,24/22,4 = 0,1 mol
=> nSO2 = 0,1 mol
<=> V SO2 = 0,1 .22,4 = 2,24 lít
b) nS = O2 = 0,1 mol
=> mS = 0,1.32 = 3,2 gam
`a)n_S=64/32=2(mol)`
`S + O_2` $\xrightarrow{t^o}$ `SO_2`
`2` `2` `(mol)`
`=>V_[O_2]=2.22,4=44,8(l)`
`b)n_[CH_4]=[4,48]/[22,4]=0,2(mol)`
`CH_4 + 2O_2` $\xrightarrow{t^o}$ `CO_2 + 2H_2 O`
`0,2` `0,4` `(mol)`
`=>V_[O_2]=0,4.22,4=8,96(l)`