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a) 2KClO3 (7/75 mol) \(\underrightarrow{t^o}\) 2KCl (7/75 mol) + 3O2\(\uparrow\) (0,14 mol).
b) Số mol khí oxi là 4,48/32=0,14 (mol).
Khối lượng kali clorat cần dùng là 7/75.122,5=343/30 (g).
Khối lượng chất rắn thu được là 7/75.74,5=1043/150 (g).
\(a,PTHH:2KClO_3\underrightarrow{t^o,MnO_2}2KCl+3O_2\uparrow\\ b,n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ Theo.pt:n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}.0,2=\dfrac{2}{15}\left(mol\right)\\ m_{KClO_3}=\dfrac{2}{15}.122,5=\dfrac{49}{3}\left(g\right)\)
\(a.n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ \Rightarrow n_{KClO_3}=0,2.\dfrac{2}{3}=\dfrac{2}{15}\left(mol\right)\\ m_{KClO_3}=\dfrac{2}{15}.122,5\approx16,333\left(g\right)\\ b.n_{KClO_3}=1,5\left(mol\right)\Rightarrow n_{O_2}=\dfrac{3}{2}.1,5=2,25\left(mol\right)\\ m_{O_2}=2,25.32=144\left(g\right)\\ c.n_{KClO_3}=0,1\left(mol\right)\\ \Rightarrow n_{KCl}=n_{KClO_3}=0,1\left(mol\right);n_{O_2}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\)
a. \(n_{O_2}=\dfrac{22.4}{22.4}=1\left(mol\right)\)
PTHH : 2KMnO4 ----to----> K2MnO4 + MnO2 + O2
2 1
\(m_{KMnO_4}=2.158=316\left(g\right)\)
b. PTHH : C + O2 ---to--->CO2
1 1 1
\(m_{CO_2}=1.44=44\left(g\right)\)
\(a.\)
\(n_{KClO_3}=\dfrac{3.675}{122.5}=0.03\left(mol\right)\)
\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)
\(0.03........................0.045\)
\(V_{O_2}=0.045\cdot22.4=1.008\left(l\right)\)
\(n_{O_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)
\(\Rightarrow n_{KClO_3}=\dfrac{0.5\cdot2}{3}=\dfrac{1}{3}mol\)
\(\Rightarrow m_{KClO_3}=\dfrac{1}{3}\cdot122.5=40.83\left(g\right)\)
a, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
b, \(n_{KMnO_4}=\dfrac{47,4}{158}=0,3\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_P=0,15\left(mol\right)\Rightarrow V_{O_2}=0,15.22,4=3,36\left(l\right)\)
c, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Theo PT: \(n_P=\dfrac{4}{5}n_{O_2}=0,12\left(mol\right)\Rightarrow m_P=0,12.31=3,72\left(g\right)\)
a) \(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: 2KClO3 --to--> 2KCl + 3O2
0,2<-------------------0,3
=> \(m_{KClO_3}=0,2.122,5=24,5\left(g\right)\)
b) \(n_{KClO_3}=\dfrac{490}{122,5}=4\left(mol\right)\)
PTHH: 2KClO3 --to--> 2KCl + 3O2
4-------------->4---->6
=> \(m_{KCl}=4.74,5=298\left(g\right)\)
=> \(m_{O_2}=6.32=192\left(g\right)\)
2KClO3 \(\underrightarrow{t^o}\) 2KCl + 3O2
a, \(n_{O_2}=\dfrac{6,72}{22,4}=0,3mol\\ n_{KClO_3}=\dfrac{0,3.2}{3}=0,2mol\\ m_{KClO_3}=0,2.122,5=24,5g\)
b, \(n_{KClO_3}=\dfrac{490}{122,5}=4mol\)
\(\Rightarrow m_{KCl}=4.74,5=298g\)
\(n_{O_2}=\dfrac{4.3}{2}=6mol\\ m_{O_2}=6.32=192g\)
a, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
b, \(n_{KClO_3}=\dfrac{19,6}{122,5}=0,16\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,24\left(mol\right)\Rightarrow V_{O_2}=0,24.22,4=5,376\left(l\right)\)
c, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Theo PT: \(n_{Al}=\dfrac{4}{3}n_{O_2}=0,32\left(mol\right)\Rightarrow m_{Al}=0,32.27=8,64\left(g\right)\)
a) \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\uparrow\)
b) số mol của 19,6 g \(KClO_3\) là:
\(n_{KClO_3}=\dfrac{m}{M}=\dfrac{19,6}{122,5}=0,16\left(mol\right)\)
thể tích của khí Oxi (đktc) là:
\(V_{O_2}=n.22,4=0,24.22,4=5,376\left(l\right)\)
c)\(PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
khối lương Al cần dùng để tác dụng hết Oxi:
\(m_{Al}=n.M=0,32.27=8,64\left(g\right)\)
PTHH: \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)
Ta có: \(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,02\left(mol\right)\\n_{Fe}=0,03\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,03\cdot56=1,68\left(g\right)\\V_{O_2}=0,02\cdot22,4=0,448\left(l\right)\end{matrix}\right.\)
a) PTHH: \(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\uparrow\left(1\right)\)
\(2KClO_3\xrightarrow[t^0]{MnO_2}2KCl+3O_2\uparrow\left(2\right)\)
b) \(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PTHH (2): \(n_{KClO_3}=\dfrac{0,3.2}{3}=0,2\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=0,2.122,5=24,5\left(g\right)\)
c) \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PTHH (1): \(n_{KMnO_4}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
\(\Rightarrow m_{KMnO_4}=0,4.158=63,2\left(g\right)\)