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\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)
\(n_{O_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(2Zn+O_2\underrightarrow{^{^{t^0}}}2ZnO\)
LTL : \(\dfrac{0.2}{2}< \dfrac{0.4}{1}\Rightarrow O_2dư\)
\(m_{O_2\left(dư\right)}=\left(0.4-0.1\right)\cdot32=9.6\left(g\right)\)
\(m_{ZnO}=0.2\cdot81=16.2\left(g\right)\)
a) \(n_{CH_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,4---------------->0,4
=> \(V_{CO_2}=0,4.22,4=8,96\left(l\right)\)
b) \(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Xét tỉ lệ \(\dfrac{0,4}{1}>\dfrac{0,4}{2}\) => CH4 dư, O2 hết
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,4-------->0,2
=> \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)
a)\(n_{Fe}=\dfrac{16,8}{56}=0,3mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,3 0,3 0,3
\(V_{H_2}=0,3\cdot22,4=6,67l\)
\(m_{FeCl_3}=0,3\cdot127=38,1g\)
b)\(n_{Fe_2O_3}=\dfrac{18}{160}=0,1125mol\)
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
0,1125 0,3 0 0
0,1 0,3 0,2 0,3
0,0125 0 0,2 0,3
\(m_{Fe}=0,2\cdot56=11,2g\)
a, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
b, \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,2}{4}< \dfrac{0,4}{5}\), ta được O2 dư.
Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=0,25\left(mol\right)\Rightarrow n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)
\(\Rightarrow m_{O_2\left(dư\right)}=0,15.32=4,8\left(g\right)\)
c, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
d, \(m_{P_2O_5}=14,2.80\%=11,36\left(g\right)\)
\(a,PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\\ n_{Al}=\dfrac{m}{M}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ Theo.PTHH:n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\\ m_{Al_2O_3}=n.M=0,2.102=20,4\left(g\right)\)
\(b,n_{O_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ Lập.tỉ.lệ:\dfrac{n_{Al}}{4}>\dfrac{n_{O_2}}{3}\Rightarrow Al.dư\\ Theo.PTHH:n_{Al\left(pư\right)}=\dfrac{4}{3}.n_{O_2}=\dfrac{4}{3}.0,2\left(mol\right)\\ n_{Al\left(dư\right)}=n_{Al\left(bđ\right)}-n_{Al\left(pư\right)}=0,4-0,2=0,2\left(mol\right)\\ Theo.PTHH:n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{Al_2O_3}=n.M=0,1=102=10,2\left(g\right)\)
\(n_P=\dfrac{6.2}{31}=0.2\left(mol\right)\)
\(n_{O_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(4P+5O_2\underrightarrow{^{^{t^0}}}2P_2O_5\)
\(4........5\)
\(0.2.........0.4\)
Lập tỉ lệ : \(\dfrac{0.2}{4}< \dfrac{0.4}{5}\Rightarrow O_2dư\)
\(n_{P_2O_5}=0.2\cdot\dfrac{2}{4}=0.1\left(mol\right)\)
\(m_{P_2O_5}=0.1\cdot142=14.2\left(g\right)\)
\(m_{P_2O_5\left(tt\right)}=14.2\cdot80\%=11.36\left(g\right)\)
a)
$4FeS_2 + 11O_2 \xrightarrow{t^o} 2Fe_2O_3 + 8SO_2$
b)
$n_{SO_2} = \dfrac{8,96}{22,4} = 0,4(mol)$
Theo PTHH :
$n_{FeS_2} = \dfrac{1}{2}n_{SO_2} = 0,2(mol)$
$m = 0,2.120 = 24(gam)$
a,\(n_{SO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: FeS2 + O2 → Fe2O3 + SO2
Mol: 0,4 0,4
\(m_{FeS_2}=0,4.120=48\left(g\right)\)