Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
2KClO3-to\xt->2KCl+3O2
0,1------------------0,1
n KClO3=\(\dfrac{12,25}{122,5}\)=0,1 mol
=>m KCl=0,1.74,5=7,45g
H=\(\dfrac{6,8}{7,45}.100\)=91,275%
b)
2KClO3-to\xt->2KCl+3O2
0,2-------------------------0,3 mol
n O2=\(\dfrac{6,72}{22,4}\)=0,3 mol
H=85%
=>m KClO3=0,2.122,5.\(\dfrac{100}{85}\)=28,82g
c)
2KClO3-to\xt->2KCl+3O2
0,2------------------------0,3
n KClO3=\(\dfrac{24,5}{122,5}\)=0,2 mol
H=80%
=>m O2=0,3.32.\(\dfrac{80}{100}\)=10,4g
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{KCl}=\dfrac{17.88}{74.5}=0.24\left(mol\right)\)
\(2KClO_3\underrightarrow{^{^{t^0}}}2KCl+3O_2\)
\(0.24.............0.24\)
\(m_{KClO_3}=\dfrac{0.24\cdot122.5}{80\%}=36.75\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{KClO_3\left(bd\right)}=\dfrac{55,125}{122,5}=0,45\left(mol\right)\)
=> \(n_{KClO_3\left(pư\right)}=\dfrac{0,45.85}{100}=0,3825\left(mol\right)\)
PTHH: 2KClO3 --to,MnO2--> 2KCl + 3O2
0,3825------------------->0,57375
=> \(V_{O_2}=0,57375.22,4=12,852\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2KClO3--t0, xt--> 2KCl + 3O2
TA có nKClO3=49/122,5=0,4
=> nKClO3 PỨ =0,4.80/100=0,32
=> nO2=3/2 . 0,32=0,48
=> mO2= 0,48.32=15,36 g
\(n_{KClO_3}=\frac{49}{122,5}=0,4mol\)
PTHH : \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo PTHH : \(n_{O_2}=\frac{3}{2}n_{KClO_3}=\frac{3}{2}0,4=0,6mol\)
=> \(m_{O_2}=0,6.32=19,2g\)
Hs = 80% ; mO2 (tt) = 19,2 / 100 .80 = 15,36 g
![](https://rs.olm.vn/images/avt/0.png?1311)
\(PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\left(1\right)\\ PTHH:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\left(2\right)\)
\(Gọi.số.mol.của.O_2.là:a\)
\(Theo.PTHH\left(1\right):n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2}{3}.a=\dfrac{2a}{3}\\ m=m_{KClO_3}=n.M=\dfrac{2a}{3}.122,5=\dfrac{245a}{3}\left(g\right)\)
\(Theo.PTHH\left(2\right):n_{KMnO_4}=2.n_{O_2}=2.a=2a\\ x=m_{KMnO_4}=n.M=2a.158=316a\left(g\right)\)
\(Tỉ.lệ:m:x:\dfrac{m}{x}=\dfrac{\dfrac{245a}{3}}{316a}=\dfrac{245}{948}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo định luật bảo toàn khối lượng, ta có khối lượng khí oxi thu được là:
m O 2 = 24,5 – 13,45 = 11,05(g)
Khối lượng thực tế oxi thu được: m O 2 = (11,05 x 80)/100 = 8,84 (g)
$2KClO_3 \xrightarrow{t^o,MnO_2} 2KCl + 3O_2$
$n_{KCl} = \dfrac{17,88}{74,5} = 0,24(mol)$
$n_{KClO_3\ pư} = 0,24 : 80\% = 0,3(mol)$
$m = 0,3.122,5 = 36,75(gam)$