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\(n_{KClO_3}=\dfrac{24,5}{122,5}=0,2mol\)
\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)
0,2 0,2 0,3 ( mol )
\(V_{O_2}=0,3.24,79=7,437l\)
\(m_{KCl}=0,2.74,5=14,9g\)
a, PTHH: 2KClO3 --to--> 2KCl + 3O2
b, \(n_{O_2}=\dfrac{53,76}{22,4}=2,4\left(mol\right)\\ n_{O_2}=2,4.32=76,8\left(g\right)\)
Bảo toàn khối lượng: \(m_{KClO_3}=76,8+168,2=245\left(g\right)\)
c, Theo pthh: \(n_{KClO_3\left(pư\right)}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}.2,4=1,6\left(mol\right)\\ \Rightarrow\%m_{KClO_3\left(phân.huỷ\right)}=\dfrac{1,6.122,5}{245}=80\%\)
a)\(n_{KClO_3}=\dfrac{36,75}{122,5}=0,3mol\)
\(2KClO_3\rightarrow2KCl+3O_2\)
0,3 0,3 0,45
\(V_{O_2}=0,45\cdot22,4=10,08l\)
b)\(n_P=\dfrac{9,3}{31}=0,3mol\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
0,3 0,45 0
0,3 0,375 0,15
0 0,075 0,15
\(m_{P_2O_5}=0,14\cdot142=19,88g\)
Gọi số mol KClO3, KMnO4 trong mỗi phần là a, b (mol)
Phần 1:
\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
mY = 122,5a + 158b - 0,1.32 = 122,5a + 158b - 3,2 (g)
Bảo toàn O: \(n_{O\left(Y\right)}=3a+4b-0,2\left(mol\right)\)
\(\%O=\dfrac{16\left(3a+4b-0,2\right)}{122,5a+158b-3,2}.100\%=34,5\%\)
=> 5,7375a + 9,49b = 2,096 (1)
Phần 2:
PTHH: 2KClO3 --to--> 2KCl + 3O2
a----------->a
2KMnO4 --to--> K2MnO4 + MnO2 + O2
b------------>0,5b------>0,5b
=> 74,5a + 142b = 29,1 (2)
(1)(2) => a = 0,2 (mol); b = 0,1 (mol)
=> \(\left\{{}\begin{matrix}\%m_{KClO_3}=\dfrac{0,2.122,5}{0,2.122,5+0,1.158}.100\%=60,8\%\\\%m_{KMnO_4}=\dfrac{0,1.158}{0,2.122,5+0,1.158}.100\%=39,2\%\end{matrix}\right.\)
\(a,PTHH:2KMnO_4\rightarrow^{t^o}K_2MnO_4+MnO_2+O_2\\ b,n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ \Rightarrow m_{O_2}=0,3\cdot32=9,6\left(g\right)\\ \Rightarrow m_{KMnO_4\left(bđ\right)}=m_{\text{chất rắn}}+m_{O_2}=109,6\left(g\right)\\ c,n_{MnO_2}=0,3\left(mol\right)\\ \Rightarrow m_{MnO_2}=0,3\cdot87=26,1\left(g\right)\\ \Rightarrow\%_{MnO_2}=\dfrac{26,1}{100}\cdot100\%=26,1\%\\ \Rightarrow\%_{KMnO_4}=100\%-26,1\%=73,9\%\)
số mol O2 là:
\(n_{O_2}=\frac{73,56}{22,4}=3,284\left(mol\right)\)
PTHH:\(2KClO_3\rightarrow2KCl+3O_2\)
\(m_{KClO_3}=n.M=\left(3,284.\frac{2}{3}\right).\left(39+35,5+16.3\right)=268,153\left(g\right)\)
\(m_{ }=n.M=\left(3,284.\frac{2}{3}\right).\left(39+35,5\right)=163,0805\left(g\right)\)(m chất rắn)
\(n_{Al}=\dfrac{1,728}{27}=0,064\left(mol\right)\)
PTHH: 4Al + 3O2 --to--> 2Al2O3
____0,064->0,048
=> mO2 = 0,048.32 = 1,536 (g)
\(m_B=\dfrac{0,894.100}{8,127}=11\left(g\right)\)
Theo ĐLBTKL: mA = mB + mO2
=> mA = 11 + 1,536 = 12,536 (g)
$2Mg + O_2 \xrightarrow{t^o} 2MgO$
$2Cu + O_2 \xrightarrow{t^o} 2CuO$
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
$MgO + 2HCl \to MgCl_2 + H_2O$
$CuO + 2HCl \to CuCl_2 + H_2O$
$Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O$
Gọi $n_{MgO} = a(mol) ; n_{CuO} = b(mol) ; n_{Al_2O_3} = c(mol)$
Bảo toàn khối lượng : $m_{O_2} = 23,2 - 16,8 = 6,4(gam)$
$n_{O_2} = 0,2(mol)$
$\Rightarrow 0,5a + 0,5b + 1,5c = 0,2(1)$
Theo PTHH :
$n_{HCl} =2 n_{MgO} + 2n_{CuO} + 6n_{Al_2O_3} = 0,8(theo (1))$
Suy ra : $V_{dd\ HCl} = \dfrac{0,8}{2} = 0,4(lít)$
a) \(n_{KCl}=\dfrac{14,9}{74,5}=0,2\left(mol\right)\)
PTHH: 2KClO3 --to--> 2KCl + 3O2
0,2<-----------0,2----->0,3
=> mKClO3 = 0,2.122,5 = 24,5(g)
VO2 = 0,3.22,4 = 6,72(l)
b) \(n_{KClO_3}=\dfrac{25,725}{122,5}=0,21\left(mol\right)\)
Gọi số mol KClO3 pư là a
=> (0,21-a).122,5 + 74,5a = 16,125
=> a = 0,2 (mol)
=> nO2 = 0,3 (mol)
=> VO2 = 0,3.22,4 = 6,72(l)
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