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a, PT: \(2Mg+O_2\underrightarrow{t^o}2MgO\)
b, Ta có: \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{Mg}=0,05\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,05.22,4=1,12\left(l\right)\)
c, PT: \(2Zn+O_2\underrightarrow{t^o}2ZnO\)
Theo PT: \(n_{ZnO}=2n_{O_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{ZnO}=0,1.81=8,1\left(g\right)\)
Bạn tham khảo nhé!
a. \(2Mg+O_2\rightarrow2MgO\)
b. \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
\(\rightarrow n_{O_2}=\dfrac{1}{2}.n_{Mg}=0,05\left(mol\right)\)
\(V_{O_2}=0,05.22,4=1,12\left(l\right)\)
c.
\(2Zn+O_2\rightarrow2ZnO\)
0,1 ...... 0,05 .... 0,1 (mol)
\(\rightarrow m_{ZnO}=0,1.81=8,1\left(g\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ n_{O_2}=\dfrac{3}{4}.0,2=0,15\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ V_{kk\left(đktc\right)}=5.3,36=16,8\left(l\right)\\ b,n_{Al_2O_3}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)
\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ n_{CuO}=n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right);n_{O_2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ a,V_{O_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ V_{kk\left(đktc\right)}=\dfrac{100.1,12}{20}=5,6\left(l\right)\\ b,m_{CuO}=0,1.80=8\left(g\right)\\ c,2R+O_2\rightarrow\left(t^o\right)2RO\\ n_R=2.n_{O_2}=2.0,05=0,1\left(mol\right)\\ M_R=\dfrac{2,4}{0,1}=24\left(\dfrac{g}{mol}\right)\\ \Rightarrow R:Magie\left(Mg=24\right)\)
\(n_K=\dfrac{m}{M}=\dfrac{7,8}{39}=0,2\left(mol\right)\)
\(a,PTHH:4K+O_2\rightarrow2K_2O\)
\(0,2:0,05:0,1\left(mol\right)\)
\(K_2O+H_2O\rightarrow2KOH\)
\(0,1:0,1:0,2\left(mol\right)\)
\(b,V_{O_2}=n.22,4=0,05.22,4=1,12\left(l\right)\)
\(c,m_{KOH}=n.M=0,2.\left(39+16+1\right)=0,2.56=11,2\left(g\right)\)
nK=mM=7,839=0,2(mol)��=��=7,839=0,2(���)
a,PTHH:4K+O2→2K2O�,����:4�+�2→2�2�
0,2:0,05:0,1(mol)0,2:0,05:0,1(���)
K2O+H2O→2KOH�2�+�2�→2���
0,1:0,1:0,2(mol)0,1:0,1:0,2(���)
b,VO2=n.22,4=0,05.22,4=1,12(l)�,��2=�.22,4=0,05.22,4=1,12(�)
c,mKOH=n.M=0,2.(39+16+1)=0,2.56=11,2(g)�,����=�.�=0,2.(39+16+1)=0,2.56=11,2(�)
a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,1-->0,2------>0,1-->0,1
=> VH2 = 0,1.22,4 = 2,24 (l)
mZnCl2 = 0,1.136 = 13,6 (g)
b) \(C\%_{dd.HCl}=\dfrac{0,2.36,5}{200}.100\%=3,65\%\)
c) \(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,15}{1}\) => H2 hết, O2 dư
PTHH: 2H2 + O2 --to--> 2H2O
0,1--------------->0,1
=> mH2O = 0,1.18 = 1,8 (g)
nAl=\(\dfrac{5,4}{27}\)=0,2 mol
2Al+3H2SO4→Al2(SO4)3+3H2
0,2-----0,3---------0,1-----------0,3
=>VH2=0,3.22,4=6,72l
=>CMH2SO4=\(\dfrac{0,3}{0,1}\)=3M
=>CM Al2(SO4)3=\(\dfrac{0,1}{0,1}\)=1M
a) S + O2 --to--> SO2
b) \(n_S=\dfrac{1,6}{32}=0,05\left(mol\right)\)
PTHH: S + O2 --to--> SO2
0,05-->0,05--->0,05
=> VSO2 = 0,05.22,4 = 1,12 (l)
c)
PTHH: 2KClO3 --to--> 2KCl + 3O2
\(\dfrac{1}{30}\)<-----------------0,05
=> \(m_{KClO_3}=\dfrac{1}{30}.122,5=\dfrac{49}{12}\left(g\right)\)
Bạn xem lời giải ở đây nhé.
https://hoc24.vn/cau-hoi/cho-324-g-al-tac-dung-voi-oxi-vua-du-th-duoc-al2o3-a-tinh-vo2-b-tinh-m-al2o3-c-trong-vkk-can-dung-biet-vo2-21-vkk-d-tinh-khoi-luong-kmno.7651142171785
\(n_P=\dfrac{6,2}{31}=0,2mol\\ 4P+5O_2\underrightarrow{t^{^{ }0}}2P_2O_5\\ n_{O_2}=\dfrac{5}{4}n_P=0,25mol\\ V_{O_2}=22,4.0,25=5,6L\\ n_S=\dfrac{3,2}{32}=0,1mol\\ S+O_2\underrightarrow{t^{^{ }0}}SO_2\\ Có:\dfrac{n_S}{1}=0,1< \dfrac{n_{O_2}}{1}=0,25\\ \Rightarrow Tính.theo.S\\ n_{SO_2}=0,1mol\\ m_{SO_2}=0,1.64=6,4g\)