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\(a.2KClO_3\xrightarrow[t^0]{xt}2KCl+3O_2\\ b.n_{KClO_3}=\dfrac{m_{KClO_3}}{M_{KClO_3}}=\dfrac{22,05}{122,5}=0,18mol\\ n_{O_2\left(lt\right)}=\dfrac{0,18.3}{2}=0,27mol\\ V_{O_2\left(lt\right)}=n_{O_2\left(lt\right)}.22,4=0,27.22,4=6,048l\\ H=\dfrac{V_{O_2\left(tt\right)}}{V_{O_2\left(lt\right)}}\cdot100\%=\dfrac{3,36}{6,048}\cdot100\%\approx55,56\%\\ c.n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{3,36}{22,4}=0,15mol\\ n_{KCl}=\dfrac{0,15.2}{3}=0,1mol\\ m_{KCl}=n_{KCl}.M_{KCl}=0,1.74,5=7,45g\)
a) PTHH: 2 KClO3 -to-> 2 KCl + 3 O2
nKCl= 14,9/74,5= 0,2(mol)
b) nKClO3=nKCl=0,2(mol)
=>mKClO3=0,2.122,5=24,5(g)
c) nO2=3/2. 0,2=0,3(mol)
=>V(O2,đktc)=0,3.22,4=6,72(l)
\(4P+5O_2\underrightarrow{^{^{t^0}}}2P_2O_5\)
Công thức khối lượng :
\(m_P+m_{O_2}=m_{P_2O_5}\)
Khi đó :
\(m_P=m_{P_2O_5}-m_P=14.2-6.2=8\left(g\right)\)
a) \(4P+5O_2\xrightarrow[]{t^o}2P_2O_5\)
b) \(m_P+m_{O_2}=m_{P_2O_5}\)
c) áp dụng định luật bảo toàn khối lượng, ta có:
\(m_P+m_{O_2}=m_{P_2O_5}\)
\(\Rightarrow m_{O_2}=m_{P_2O_5}-m_P=14,2-6,2=8\left(g\right)\)
vậy khối lượng oxi đã phản ứng là \(8g\)
2)
nH2 = 6.72/22.4 = 0.3 (mol)
Fe2O3 + 3H2 -to-> 2Fe + 3H2O
0.1______0.3______0.2
mFe2O3 = 0.1*160 = 16 (g)
mFe = 0.2*56 = 11.2 (g)
3)
nFe3O4 = 11.6/232 = 0.05 (mol)
3Fe + 2O2 -to-> Fe3O4
0.15___0.1______0.05
mFe = 0.15*56 = 8.4 (g)
VO2 = 0.1*22.4 = 2.24 (l)
2KClO3 -to-> 2KCl + 3O2
1/15______________0.1
mKClO3 = 1/15 * 122.5 = 8.167 (g)
a)
3H2 + Fe2O3 --to--> 2Fe + 3H2O
b) nH2 = 6,72/22,4 = 0,3 mol
Từ pt => nFe3O4 = 0,1 mol
=> mFe3O4 = 0,1. 232 = 23,2 g
a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)=n_{MgCl_2}\)
\(\Rightarrow m_{MgCl_2}=0,15\cdot95=14,25\left(g\right)\)
c) Theo PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(p.ứ\right)}=0,3\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl\left(p.ứ\right)}=0,3\cdot36,5=10,95\left(g\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Mg}+m_{ddHCl}-m_{H_2}=53,3\left(g\right)\)
\(\Rightarrow m_{ddHCl}=50\left(g\right)\) \(\Rightarrow C\%_{HCl\left(p.ứ\right)}=\dfrac{10,95}{50}\cdot100\%=21,9\%\)
a, \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,1\left(mol\right)\Rightarrow m_{Fe_2O_3}=0,1.160=16\left(g\right)\)
c, Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Fe}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
`2KClO_3->2KCl+3O_2`(to)
0,04-----------0,02-----0,06
`n_(KClO_3)=(4,9)/(122,5)=0,04mol`
=>`V_(O_2)=0,06.24,79=1,4847l`
c)
`4P+5O_2->2P_2O_5`(to)
0,048----0,06 mol
`=>m_P=0,048.31=1,488g`
\(a,PTHH:2KClO_3\xrightarrow{t^o}2KCl+3O_2\\ b,n_{KClO_3(thực tế)}=\dfrac{36,75}{122,5}=0,3(mol)\\ n_{O_2(phản ứng)}=\dfrac{6,72}{22,4}=0,3(mol)\\ \Rightarrow n_{KClO_3(phản ứng)}=\dfrac{2}{3}n_{O_2}=0,2(mol)\\ \Rightarrow H\%=\dfrac{0,2}{0,3}.100\%=66,67\%\\ c,n_{KCl}=n_{KClO_3(phản ứng)}=0,2(mol)\\ \Rightarrow m_{KCl}=0,2.74,5=14,9(g)\)
a)PTHH:\(2KCLO_3\rightarrow^{t^o}2KCL+3O_2\)
b)\(n_{KCL}=\frac{2,98}{74,5}=0,4mol\)
Từ phương trình hóa học\(n_{KCL}=n_{KCLO_3}=0,04mol\)
\(n_{KCLO_3}=0,04\cdot122,5=4,9g\)
\(H=\frac{4,9}{9,8}\cdot100\%=50\%\)
a. PTHH: \(2KClO_3\rightarrow^{t^o}2KCl+3O_2\)
b. \(n_{KCl}=\frac{2,98}{74,5}=0,4mol\)
Từ phương trình hoá học \(n_{KCl}=n_{KClO_3}=0,04mol\)
\(n_{KClO_3}=0,04.122,5=4,9g\)
\(H=\frac{4,9}{9,8}.100\%=50\%\)