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\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\\
pthh:2K+2H_2O\rightarrow2KOH+H_2\uparrow\)
QT chuyển xanh
\(pthh:2K+2H_2O\rightarrow2KOH+H_2\)
0,2 0,2 0,1
\(V_{H_2}=0,1.22,4=2,24\left(L\right)\\
m_{KOH}=0,2.56=11,2\left(g\right)\)
\(pthh:Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
0,1 0,075
=> \(m_{Fe}=\left(0,075.56\right).80\%=3,36g\)
\(a) 4Na + O_2 \xrightarrow{t^o} 2Na_2O\\ b) n_{Na} = \dfrac{4,6}{23} = 0,2(mol)\\ n_{O_2} = \dfrac{1}{4}n_{Na} = 0,05(mol)\\ V_{O_2} = 0,05.22,4 = 1,12(lít)\\ c) Na_2O + H_2O \to 2NaOH\\ n_{NaOH} = n_{Na} = 0,2(mol)\\ C\%_{NaOH} = \dfrac{0,2.40}{160}.100\% = 5\%\\ d)\)
\(n_{Na\ thêm} = x(mol)\\ 2Na + 2H_2O \to 2NaOH + H_2\\ n_{NaOH} = n_{Na} = x(mol)\\ n_{H_2} =0,5x(mol)\\ \Rightarrow m_{dd} = 23x + 160 -0,5x.2 = 22x + 160(gam)\\ \Rightarrow C\% = \dfrac{0,2.40 + 40x}{22x + 160}.100\% = 5\% + 5\%\\ \Rightarrow x = \dfrac{40}{189}\\ m_{Na} = \dfrac{40}{189}.23 = 4,87(gam)\)
\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)
\(4Na+O_2\underrightarrow{^{t^0}}2Na_2O\)
\(0.2.....0.05.........0.1\)
\(V_{O_2}=0.05\cdot22.4=1.12\left(l\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
\(0.1.......................0.2\)
\(m_{NaOH}=0.2\cdot40=8\left(g\right)\)
\(C\%_{NaOH}=\dfrac{8}{160}\cdot100\%=5\%\)
Để C% tăng thêm 5%
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
\(a...............a.......0.5a\)
\(m_{NaOH}=40a\left(g\right)\)
\(m_{dd_{NaOH}}=23a+160-0.5a\cdot2=22a+160\left(g\right)\)
\(C\%_{NaOH}=\dfrac{40a+8}{22a+160}\cdot100\%=5\%\)
\(\Rightarrow a=0\)
=> Sai đề
a, PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Ta có: \(n_{Al}=\dfrac{10,2}{27}=\dfrac{17}{45}\left(mol\right)\)
b, Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=\dfrac{17}{60}\left(mol\right)\)
\(\Rightarrow V_{O_2}=\dfrac{17}{60}.22,4\approx6,347\left(l\right)\)
c, Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac{17}{90}\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=\dfrac{17}{90}.102\approx19,267\left(g\right)\)
d, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=\dfrac{17}{30}\left(mol\right)\)
\(\Rightarrow m_{KMnO_3}=\dfrac{17}{30}.158\approx89,53\left(g\right)\)
a)PTHH:2KClO\(_3\)➞\(^{t^o}\)2KCl+3O\(_2\)
b) n\(_{KClO_3}\)=\(\dfrac{m_{KClO_3}}{M_{KClO_3}}\)=\(\dfrac{12,15}{122,5}\)\(\approx\)0,1(m)
PTHH : 2KClO\(_3\) ➞\(^{t^o}\) 2KCl + 3O\(_2\)
tỉ lệ : 2 2 3
số mol : 0,1 0,1 0,15
V\(_{O_2}\)=n\(_{O_2}\).22,4=0,15.22,4=3,36(l)
c)PTHH : 2Zn + O\(_2\) -> 2ZnO
tỉ lệ : 2 1 2
số mol :0,3 0,15 0,3
m\(_{Zn}\)=n\(_{Zn}\).M\(_{Zn}\)=0,3.65=19,5(g)
2K + O2 → 2K2O
nK2O = \(\dfrac{18,8}{94}\)= 0,2 mol => nKphản ứng = 0,2 mol , nO2phản ứng = 0,1 mol
VO2 = 0,1.22,4 = 2,24 lít . Oxi chiếm 1/5 thể tích không khí => V không khí = 2,24.5 = 11,2 lít
mK = 0,2.39 = 7,8 gam
2K + O2 → 2K2O
Nếu có 3,36 lít Oxi phản ứng với 0,2 mol kali => nO2 = \(\dfrac{3,36}{22,4}\)= 0,15mol
Ta có tỉ lệ \(\dfrac{0,2}{2}< \dfrac{0,15}{1}\)=> Oxi dư , kali hết .
Khối lượng sp thu được vẫn tính theo kali => nK2O = 0,2 mol
<=> mK2O = 0,2.94 = 18,8 gam
\(n_P=\dfrac{m}{M}=\dfrac{12,4}{31}=0,4\left(mol\right)\\ PTHH:4P+5O_2-^{t^o}>2P_2O_5\)
Tỉ lệ 4 : 5 : 2
n(mol) 0,4--->0,5----->0,2
\(V_{O_2\left(dktc\right)}=n\cdot22,4=0,5\cdot22,4=11,2\left(l\right)\\ V_{kk}=11,2:\dfrac{1}{5}=56\left(l\right)\)
\(m_{P_2O_5}=n\cdot M=0,2\cdot142=28,4\left(g\right)\)
\(PTHH:P_2O_5+3H_2O->2H_3PO_4\)
tỉ lệ 1 : 3 : 2
n(mol) 0,2----->0,6--------->0,4
\(m_{H_3PO_4}=n\cdot M=0,4\cdot98=39,2\left(g\right)\)
\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{39,2}{200}\cdot100\%=19,6\%\)
Bài 1:
a, \(S+O_2\underrightarrow{t^o}SO_2\)
b, Ta có: \(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)
Theo PT: \(n_{SO_2}=n_S=0,1\left(mol\right)\Rightarrow m_{SO_2}=0,1.64=6,4\left(g\right)\)
c, \(n_{O_2}=n_S=0,1\left(mol\right)\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)
Bài 2:
a, \(2KClO_3\xrightarrow[MnO_2]{^{t^o}}2KCl+3O_2\)
b, Bạn xem lại đề nhé, pư không tạo thành MnO2.
Bài 3:
a, \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{Cu}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)
c, \(n_{H_2O}=n_{H_2}=0,1\left(mol\right)\Rightarrow V_{H_2O}=0,1.22,4=2,24\left(l\right)\)
d, \(n_{CuO}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{CuO}=0,1.80=8\left(g\right)\)
nK = 7,8 : 39 = 0,2 (mol)
pthh : 2K + 2H2O ---> 2KOH + H2
0,2--------------------------> 0,1 (mol)
=> VO2 = 0,1 . 22,4 = 2,24 (l)
\(n_K=\dfrac{m}{M}=\dfrac{7,8}{39}=0,2\left(mol\right)\)
\(a,PTHH:4K+O_2\rightarrow2K_2O\)
\(0,2:0,05:0,1\left(mol\right)\)
\(K_2O+H_2O\rightarrow2KOH\)
\(0,1:0,1:0,2\left(mol\right)\)
\(b,V_{O_2}=n.22,4=0,05.22,4=1,12\left(l\right)\)
\(c,m_{KOH}=n.M=0,2.\left(39+16+1\right)=0,2.56=11,2\left(g\right)\)
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0,2:0,05:0,1(mol)0,2:0,05:0,1(���)
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