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2KMnO4-to>K2MnO4+MnO2+O2
0,3-----------------0,15-----0,15------0,15 mol
n KMnO4=\(\dfrac{47,4}{158}\)=0,3 mol
=>mcr=0,15.197.0,15.87=42,6g
=>VO2=0,15.22,4=3,36l
b) 4P+5O2-to>2P2O5
0,1--------------0,05
nP=\(\dfrac{3,1}{31}\)=0,1 mol
->O2 dư
=>m P2O5=0,05.142=7,1g
mKMnO4 = 47,4/158 = 0,3 (mol)
PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2
Mol: 0,3 ---> 0,15 ---> 0,15 ---> 0,15
m = 0,15 . 197 + 0,15 . 87 = 85,2 (g)
V = VO2 = 0,15 . 22,4 = 3,36 (l)
nP = 3,1/31 = 0,1 (mol)
PTHH: 4P + 5O2 -> (t°) 2P2O5
LTL: 0,1/4 < 0,15/5 => O2 dư
nP2O5 = 0,1/2 = 0,05 (mol)
mP2O5 = 0,05 . 142 = 7,1 (g)
a)
\(\left\{{}\begin{matrix}n_{C_2H_2}+n_{CH_4}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\\dfrac{n_{C_2H_2}}{n_{CH_4}}=2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{C_2H_2}=0,4\left(mol\right)\\n_{CH_4}=0,2\left(mol\right)\end{matrix}\right.\)
PTHH: 2C2H2 + 5O2 --to--> 4CO2 + 2H2O
0,4----->1------------->0,8
CH4 + 2O2 --to--> CO2 + 2H2O
0,2-->0,4---------->0,2
=> VO2 = (1+0,4).22,4 = 31,36(l)
=> VCO2 = (0,8 + 0,2).22,4 = 22,4 (l)
b)
PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
2,8<---------------------------------1,4
=> \(m_{KMnO_4\left(PTHH\right)}=2,8.158=442,4\left(g\right)\)
=> mKMnO4 (thực tế) = 442,4 : 80% = 553(g)
a, Theo giả thiết ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
\(4P+5O_2--t^o->2P_2O_5\)
Ta có: \(n_{O_2}=\dfrac{5}{4}.n_P=0,125\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=0,125.22,4=2,8\left(l\right)\)
b, Theo giả thiết ta có: \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(CH_4+2O_2--t^o->CO_2+2H_2O\)
Ta có: \(n_{O_2}=2.n_{CH_4}=0,1\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=2,24\left(l\right)\)
Câu 2:
\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)
0,6<------------------------------------0,3
\(\Rightarrow m_{KMnO_4}=0,6.158=94,8\left(g\right)\)
Câu 3:
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2<-----------------------0,2
=> mZn = 0,2.65 = 13 (g)
Câu 4:
\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,4------------------------->0,4
\(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
0,4<---0,4
\(\Rightarrow m_{CuO}=0,4.80=32\left(g\right)\)
\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25mol\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,25 0,5 0,25 ( mol )
\(m_{CH_4}=0,25.16=4g\)
\(V_{O_2}=0,5.22,4=11,2l\)
\(n_{O_2\left(đktc\right)}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ 4P+5O_2\underrightarrow{^{to}}2P_2O_5\\ 0,12........0,15.........0,06\left(mol\right)\\ m_P=0,12.31=3,72\left(g\right)\)
Fe+2HCl->FeCl2+H2
0,1---------------------0,1
2H2+O2-to>2H2O
0,1----0,05 mol
0,1--0,1
n Fe=\(\dfrac{5,6}{56}\)=0,1 mol
=>VH2=0,1.22,4=2,24l
=>mkk=0,05.29=1,45l
\(n_P=\dfrac{7,44}{31}=0,24mol\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
0,24 0,3 0,12
\(V_{O_2}=0,3\cdot22,4=6,72l\)
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
0,2 0,3
\(m_{KClO_3}=0,2\cdot122,5=24,5g\)
a) $C_4H_{10} + \dfrac{13}{2}O_2 \xrightarrow{t^o} 4CO_2 + 5H_2O$
b) Theo PTHH : $V_{O_2} = \dfrac{13}{2}V_{C_4H_{10}} = 21,84(lít)$
$n_{C_4H_{10}} = 0,15(mol) \Rightarrow n_{H_2O} = 5n_{C_4H_{10}} = 0,75(mol)$
$m = 0,75.18 = 13,5(gam)$
c) $2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2$
$n_{KMnO_4\ pư} = 2n_{O_2} = 1,95(mol)$
$n_{KMnO_4\ đã\ dùng} = 1,95 : 80\% = 2,4375(mol)$
$m_{KMnO_4} = 2,4375.158 = 385,125(gam)$
\(\begin{array}{l}
a)\\
2{C_4}{H_{10}} + 13{O_2} \xrightarrow{t^0} 8C{O_2} + 10{H_2}O\\
b)\\
{n_{{C_4}{H_{10}}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{{O_2}}} = 0,15 \times \dfrac{{13}}{2} = 0,975\,mol\\
{V_{{O_2}}} = 0,975 \times 22,4 = 21,84l\\
{m_{{O_2}}} = 0,975 \times 32 = 31,2g\\
c)\\
{n_{KMn{O_4}}} = 2{n_{{O_2}}} = 1,95\,mol\\
{m_{KMn{O_4}}} = \dfrac{{1,95 \times 158}}{{80\% }} = 385,125g
\end{array}\)