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Bài 6:
a) \(n_{O_2\left(tt\right)}=\dfrac{2,88}{24}=0,12\left(mol\right)\)
=> \(n_{O_2\left(PTHH\right)}=\dfrac{0,12.100}{80}=0,15\left(mol\right)\)
PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
0,3<--------------------------------0,15
=> \(m_{KMnO_4}=0,3.158=47,4\left(g\right)\)
b) \(n_{KClO_3}=\dfrac{2,45}{122,5}=0,02\left(mol\right)\)
\(V_{O_2\left(tt\right)}=8.0,072=0,576\left(l\right)\)
=> \(n_{O_2\left(tt\right)}=\dfrac{0,576}{24}=0,024\left(mol\right)\)
PTHH: 2KClO3 --to--> 2KCl + 3O2
0,02---------------->0,03
=> nO2(hao hụt) = 0,03 - 0,024 = 0,006 (mol)
=> %O2 bị hao hụt = \(\dfrac{0,006}{0,03}.100\%=20\%\)
144cm^3 = 0,114 l
Số lượng oxi cần đun:
\(V_{O_2}=\frac{0,144}{40\%}=0,36\left(l\right)\)
\(n_{O_2}=\frac{V_{O_2}}{22,4}=\frac{0,36}{22,4}=0,016\left(mol\right)\)
\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(0,032\) \(0,016\) \(0,016\) \(0,016\) \(\left(mol\right)\)
\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=0,032.158=5,056\left(g\right)\)
nO2= \(\frac{0,144}{22,4}\)=\(\frac{9}{1400}\) mol
Vì O2 thu đc chỉ chiếm 60% lượng tạo thành nên thực tế đã tạo ra 3/280 mol O2
2KMnO4 \(\underrightarrow{^{to}}\) K2MnO4+ MnO2+ O2
\(\rightarrow\) nKMnO4=\(\frac{3}{140}\)mol
\(\rightarrow\)mKMnO4= 3,39g
nO2 = \(\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
pt: \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
Theo pt: \(n_{KMnO_4}=2n_{O_2}=0,6mol\)
=> nKMnO4 thực tế = 0,6:\(\dfrac{90}{100}=\dfrac{2}{3}\left(mol\right)\)
mKMnO4 = \(\dfrac{2}{3}.158=\dfrac{316}{3}g\)
Câu 8:
\(d_{\dfrac{A}{KK}}>1\\ \Leftrightarrow M_A>M_{KK}\\ \Leftrightarrow M_A>29\\ Vậy:Chọn.A\)
(Vì 44>29>28>2)
\(Câu.7:C\\ Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\\ Câu.6:A\)
$a) CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
b) $n_{CH_4} = \dfrac{3,92}{22,4} = 0,175(mol)$
$n_{O_2} = \dfrac{3,84}{32} = 0,12(mol)$
Ta thấy : $n_{CH_4} : 1 > n_{O_2} : 2$ nên $CH_4$ dư
$n_{CH_4\ pư} = \dfrac{1}{2}n_{O_2} = 0,06(mol)$
$\Rightarrow m_{CH_4\ dư} = (0,175 - 0,06).16 = 1,84(gam)$
c) $2NaOH + CO_2 \to Na_2CO_3 + H_2O$
Theo PTHH :
$n_{Na_2CO_3} = n_{CO_2} = \dfrac{1}{2}n_{CH_4} = 0,06(mol)$
$m_{Na_2CO_3} = 0,06.106 = 6,36(gam)$
nFe = 16.8/56 = 0.3 (mol)
nO2 = 6.72/22.4 = 0.3 (mol)
2Fe + 3O2 -to-> Fe3O4
0.2___0.3________0.1
mFe dư = ( 0.3 - 0.2 ) * 56 = 5.6 (g)
mFe3O4 = 0.1*232 = 23.2 (g)
a)
3Fe+2O2→Fe3O4
b)
nFe=16,8/56=0,3mol
nO2=6,72/22,4=0,3mol
Ta có: 0,3/3<0,3/2=> O2 dư tính theo Fe
nFe3O4=0,3/3=0,1
mFe3O4=0,1.232=23,2g
PTHH: \(Cu_2S+2O_2\xrightarrow[]{t^o}2CuO+SO_2\)
a) Ta có: \(n_{Cu_2S}=\dfrac{100}{160}=0,625\left(mol\right)\) \(\Rightarrow n_{O_2\left(lýthuyết\right)}=1,25\left(mol\right)\)
\(\Rightarrow V_{O_2\left(thực\right)}=\dfrac{1,25\cdot22,4}{96\%}\approx29,17\left(l\right)\)
b) Sửa đề: "Tính khối lượng KMnO4 để hấp thụ hết SO2"
PTHH: \(5SO_2+2KMnO_4+2H_2O\rightarrow K_2SO_4+2MnSO_4+2H_2SO_4\)
Ta có: \(n_{SO_2\left(thực\right)}=n_{Cu_2S}\cdot96\%=0,6\left(mol\right)\)
\(\Rightarrow n_{KMnO_4}=0,24\left(mol\right)\) \(\Rightarrow m_{KMnO_4}=0,24\cdot158=37,92\left(g\right)\)
c) PTHH: \(SO_2+\dfrac{1}{2}O_2\xrightarrow[V_2O_5]{t^o}SO_3\)
Theo PTHH: \(n_{O_2}=\dfrac{1}{2}n_{SO_2}=0,3\left(mol\right)\) \(\Rightarrow V_{kk}=\dfrac{0,3\cdot22,4}{21\%}=32\left(l\right)\)
d) Bảo toàn nguyên tố Lưu huỳnh: \(n_{H_2SO_4\left(lýthuyết\right)}=n_{SO_2\left(thực\right)}=0,3\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(thực\right)}=0,3\cdot85\%=0,255\left(mol\right)\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,255\cdot98}{10\%}=249,9\left(g\right)\)
a, Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
Xét tỉ lệ: \(\dfrac{0,3}{2}< \dfrac{0,2}{1}\), ta được O2 dư.
Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow n_{O_2\left(dư\right)}=0,05\left(mol\right)\Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)
b, \(n_{H_2O}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{H_2O}=0,3.18=5,4\left(g\right)\)
c, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
_______0,3_______________________0,15 (mol)
\(\Rightarrow m_{KMnO_4}=0,3.158=47,4\left(g\right)\)
Bạn tham khảo nhé!
Câu 2:
\(4Al+3O2-->2Al2O3\)
a)\(n_{Al}=\frac{5,4}{27}=0,2\left(mol\right)\)
n\(_{O2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_{Al}\left(\frac{0,2}{4}\right)< nO2\left(\frac{0,3}{3}\right)\Rightarrow O2dư\)
n\(_{O2}=\frac{3}{4}n_{Al}=0,15\left(mol\right)\)
\(n_{O2}dư=0,3-0,15=0,15\left(mol\right)\)
m\(_{O2}dư=0,15.32=4,8\left(g\right)\)
b) \(n_{Al2O3}=\frac{1}{2}n_{Al}=0,15\left(mol\right)\)
\(m_{Al2O3}=0,15.102=15,3\left(g\right)\)