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a, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Ta có: \(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\)
Theo PT: \(n_{K_2MnO_4}=\dfrac{1}{2}n_{KMnO_4}=0,1\left(mol\right)\)
\(\Rightarrow m_{K_2MnO_4}=0,1.197=19,7\left(g\right)\)
b, Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,1\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,1.24,79=2,479\left(l\right)\)
c, PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Theo PT: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{1}{2}n_{O_2}=0,05\left(mol\right)\\n_{H_2O}=n_{O_2}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow V_{CO_2}=0,05.24,79=1,2395\left(l\right)\)
\(m_{H_2O}=0,1.18=1,8\left(g\right)\)
Ta có: \(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\)
PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{K_2MnO_4}=n_{MnO_2}=n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,1\left(mol\right)\)
\(\Rightarrow m_A=m_{K_2MnO_4}+m_{MnO_2}=0,1.197+0,1.87=28,4\left(g\right)\)
Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
Xét tỉ lệ: \(\dfrac{0,2}{3}>\dfrac{0,1}{2}\), ta được Fe dư.
Chất rắn B gồm: Fe3O4 và Fe dư.
⇒ mB = mFe3O4 + mFe (dư) = mFe + mO2 = 11,2 + 0,1.32 = 14,4 (g)
a, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
b, Ta có: \(n_{KMnO_4}=\dfrac{3,16}{158}=0,02\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,01\left(mol\right)\)
\(\Rightarrow m_{O_2}=0,01.32=0,32\left(g\right)\)
c, \(V_{O_2}=0,01.24,79=0,2479\left(l\right)\)
Câu 6.
\(n_{O_2}=\dfrac{16,8}{22,4}=0,75mol\)
\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
1,5 0,75
\(m_{KMnO_4}=1,5\cdot158=237g\)
Câu 7.
\(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02mol\)
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
0,04 0,02
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
\(\dfrac{2}{75}\) 0,04
\(m_{KClO_3}=\dfrac{2}{75}\cdot122,5=\dfrac{49}{15}\approx3,27g\)
\(a,n_{CuO}=\dfrac{59}{80}=0,7375\left(mol\right)\\ PTHH:2Cu\left(NO_3\right)_2\rightarrow^{t^o}2CuO+4NO_2\uparrow+O_2\uparrow\\ \Rightarrow\left\{{}\begin{matrix}n_{O_2}=\dfrac{1}{2}n_{CuO}=0,36875\left(mol\right)\\n_{NO_2}=2n_{CuO}=1,475\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,36875\cdot22,4=8,26\left(l\right)\\V_{NO_2}=1,475\cdot22,4=33,04\left(l\right)\end{matrix}\right.\)
\(b,\text{Chất rắn thu đc là }CuO\text{ gồm có }Cu,O\\ \%_O=\dfrac{16}{80}\cdot100\%=20\%\\ \Rightarrow m_O=59\cdot20\%=11,8\left(g\right)\\ \Rightarrow m_{Cu}=59-11,8=47,2\left(g\right)\)
a) nAl=2,7/27=0,1(mol)
PTHH: 2Al + 6 HCl -> 2 AlCl3 + 3H2
0,1_________0,3___0,1_____0,15(mol)
b) mHCl=0,3.36,5=10,95(g)
c) mAlCl3=0,1.133,5=13,35(g)
d) V(H2,đktc)=0,15.22,4=3,36(l)
\(n_{KMnO_4}=\dfrac{118,5}{158}=0,75\left(mol\right)\\
pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,75 0,375
=> \(V_{O_2}=0,375.22,4=8,4\left(l\right)\\
V_{kk}=8,4.5=42\left(l\right)\)
\(a,PTHH:2KMnO_4\rightarrow^{t^o}K_2MnO_4+MnO_2+O_2\\ b,n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ \Rightarrow m_{O_2}=0,3\cdot32=9,6\left(g\right)\\ \Rightarrow m_{KMnO_4\left(bđ\right)}=m_{\text{chất rắn}}+m_{O_2}=109,6\left(g\right)\\ c,n_{MnO_2}=0,3\left(mol\right)\\ \Rightarrow m_{MnO_2}=0,3\cdot87=26,1\left(g\right)\\ \Rightarrow\%_{MnO_2}=\dfrac{26,1}{100}\cdot100\%=26,1\%\\ \Rightarrow\%_{KMnO_4}=100\%-26,1\%=73,9\%\)
a)
\(2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\)
b)
\(n_{KMnO_4} = \dfrac{1,58}{158} = 0,01(mol)\\ \Rightarrow n_{O_2} = \dfrac{1}{2}n_{KMnO_4} = 0,005(mol)\\ \Rightarrow m_{K_2MnO_4} + m_{MnO_2} = m_{KMnO_4} - m_{O_2} = 1,58-0,005.32 = 1,42(gam)\)
c)
\(V_{O_2} = 0,005.22,4 = 0,112(lít)\)