Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{O_2}=\dfrac{9.6}{32}=0.3\left(mol\right)\)
\(2KClO_3\underrightarrow{^{^{t^0}}}2KCl+3O_2\)
\(a.\)
\(n_{KClO_3}=n_{KCl}=\dfrac{2}{3}\cdot n_{O_2}=\dfrac{2}{3}\cdot0.3=0.2\left(mol\right)\)
\(m_{KClO_3}=0.2\cdot122.5=24.5\left(g\right)\)
\(b.\)
\(m_{KCl}=0.2\cdot74.5=14.9\left(g\right)\)
\(a,n_{O_2}=\dfrac{9,6}{32}=0,3(mol)\\ 2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\\ \Rightarrow n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,2(mol)\\ \Rightarrow m_{KClO_3}=0,2.122,5=24,5(g)\\ b,n_{KCl}=n_{KClO_3}=0,2(mol)\\ \Rightarrow m_{KCl}=0,2.74,5=14,9(g)\)
2KClO3 -> 2KCl + 3O2
a.nO2 = 0.28125mol
=> nKClO3 = 0.1875mol
=> mKClO3 = 22.97g
b.nKCl = nKClO3 = 0.1875mol
=> mKCl = 13.97g
$a)PTHH:2KClO_3\xrightarrow{t^o}2KCl+3O_2$
$n_{O_2}=\dfrac{9}{32}=0,28125(mol)$
$\Rightarrow n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,1875(mol)$
$\Rightarrow m_{KClO_3}=0,1875.122,5=22,96875(g)$
$b)$ Theo PT: $n_{KCl}=n_{KClO_3}=0,1875(mol)$
$\Rightarrow m_{KCl}=0,1875.74,5=13,96875(g)$
\(n_{O_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ PTHH:2KClO_3-^{t^o}>2KCl+3O_2\)
tỉ lệ 2 : 2 : 3
n(mol) `1/3`<------------`1/3`<-----`0,5`
\(m_{KClO_3}=n\cdot M=\dfrac{1}{3}\cdot\left(39+35,5+16\cdot3\right)\approx40,83\left(g\right)\)
\(a,PTHH:2KClO_3\xrightarrow{t^o}2KCl+3O_2\\ b,n_{KClO_3(thực tế)}=\dfrac{36,75}{122,5}=0,3(mol)\\ n_{O_2(phản ứng)}=\dfrac{6,72}{22,4}=0,3(mol)\\ \Rightarrow n_{KClO_3(phản ứng)}=\dfrac{2}{3}n_{O_2}=0,2(mol)\\ \Rightarrow H\%=\dfrac{0,2}{0,3}.100\%=66,67\%\\ c,n_{KCl}=n_{KClO_3(phản ứng)}=0,2(mol)\\ \Rightarrow m_{KCl}=0,2.74,5=14,9(g)\)
PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Ta có: \(n_{Al_2O_3}=\dfrac{15,3}{102}=0,15\left(mol\right)\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}.\dfrac{3}{2}n_{Al_2O_3}=0,15\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=0,15.122,5=18,375\left(g\right)\)
a) \(n_{Fe}=\dfrac{33,6}{56}=0,6\left(mol\right)\)
PTHH: \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)
0,6--->0,4------->0,2 (mol)
=> \(m_{Fe_3O_4}=0,2.232=46,4\left(g\right)\)
b) \(V_{O_2\left(\text{đ}kc\right)}=0,4.24,79=9,916\left(l\right)\)
c) PTHH: \(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\)
\(\dfrac{4}{15}\)<-------------------0,4 (mol)
=> \(m_{KClO_3}=\dfrac{4}{15}.122,5=\dfrac{98}{3}\left(g\right)\)
a) \(n_{KClO_3}=\dfrac{36,75}{122,5}=0,3\left(mol\right)\)
PTHH: 2KClO3 --to--> 2KCl + 3O2
0,3---------->0,3--->0,45
b) mKCl = 0,3.74,5 = 22,35 (g)
mO2 = 0,45.32 = 14,4 (g)
VO2 = 0,45.22,4 = 10,08 (l)
c)
PTHH: 4R + nO2 --to--> 2R2On
\(\dfrac{1,8}{n}\)<-0,45
=> \(M_R=\dfrac{5,4}{\dfrac{1,8}{n}}=3n\left(g/mol\right)\)
Xét n = 4 thỏa mãn => MR = 12 (g/mol)
=> R là C (Cacbon)
\(a.2KClO_3-^{t^o}\rightarrow2KCl+3O_2\\ n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,6\left(mol\right)\\ \Rightarrow V_{O_2}=0,6.22,4=13,44\left(l\right)\\ n_{KCl}=n_{KClO_3}=0,4\left(mol\right)\\ \Rightarrow m_{KCl}=0,4.74,5=29,8\left(g\right)\)
a/ PTHH: 2KClO3 =(nhiệt)=> 2KCl + 3O2
nO2 = 9,6 / 32 = 0,3 mol
=> nKClO3 = 0,2 (mol)
=> mKClO3 = 0,2 x 122,5 = 24,5 gam
b/ Cách 1:
nKCl = nKClO3 = 0,2 mol
=> mKCl = 0,2 x 74,5 = 14,9 gam
Cách 2:
Áp dụng định luật bảo toàn khối lượng
=> mKCl = mKClO3 - mO2 = 24,5 - 9,6 = 14,9 gam