Đốt a gam C và thu được 2,9748 lít khí CO2 ở ĐKC , tính giá trị a biết chỉ có 80% C đã tham gia vào phản ứng hóa học
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\(S+O_2\underrightarrow{t^o}SO_2\)
\(1:1:1:1\)
\(0,2:0,2:0,2:0,2\left(mol\right)\)
\(n_{SO_2}=\dfrac{V}{24,79}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)
\(a,m_S=n.M=0,2.32=6,4\left(g\right)\)
\(b,V_{O_2}=n.24,79=0,2.24,79=4,958\left(l\right)\)
làm lại ko để ý có điều kiện=))))
\(n_{SO_2\left(dkc\right)}=\dfrac{V}{24,79}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)
\(PTHH:S+O_2-^{t^o}>SO_2\)
tỉ lệ 1 : 1 : 1
n(mol) 0,2<--0,2<---0,2
\(m_S=n\cdot M=0,2\cdot32=6,4\left(g\right)\\ V_{O_2\left(dkc\right)}=n\cdot24,79=0,2\cdot24,79=4,958\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2Na + H2SO4 --> Na2SO4 + H2
b) \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
PTHH: 2Na + H2SO4 --> Na2SO4 + H2
_____0,2------>0,1-------------------->0,1
=> VH2 = 0,1.22,4 = 2,24 (l)
c) mH2SO4 = 0,1.98 = 9,8(g)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)2Al + 6CH_3COOH \to 2(CH_3COO)_3Al + 3H_2\\ b)n_{Al} = \dfrac{2,7}{27} = 0,1(mol) ; n_{CH_3COOH} = \dfrac{200.10\%}{60} = \dfrac{1}{3}(mol)\\ n_{CH_3COOH} = \dfrac{1}{3}> 3n_{Al} = 0,3 \to CH_3COOH\ dư\\ n_{H_2} = \dfrac{3}{2}n_{Al} = 0,15(mol) \Rightarrow V_{H_2} = 0,15.22,4 = 3,36(lít)\\ n_{CH_3COOH\ pư} = 3n_{Al} =0,3(mol) \Rightarrow m_{CH_3COOH\ pư} = 0,3.60 = 18(gam)\\ c) m_{dd} = 2,7 + 200 - 0,15.2 = 202,4(gam)\\ n_{(CH_3COO)_3Al} = n_{Al} = 0,1(mol)\\ m_{CH_3COOH\ dư} = 200.10\% - 18 = 2(gam)\\ C\%_{(CH_3COO)_3Al} = \dfrac{0,1.204}{202,4}.100\% = `10,08\%\\ \)
\(C\%_{CH_3COOH} = \dfrac{2}{202,4}.100\% = 0,988\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Mg + H2SO4 --> MgSO4 + H2
b) \(n_{Mg}=\dfrac{14,4}{24}=0,6\left(mol\right)\)
PTHH: Mg + H2SO4 --> MgSO4 + H2
0,6--->0,6------->0,6----->0,6
=> \(m_{H_2SO_4}=0,6.98=58,8\left(g\right)\)
c)
PTHH: 2H2 + O2 --to--> 2H2O
0,6-->0,3
=> VO2 = 0,3.24,79 = 7,437 (l)
=> Vkk = 7,437.5 = 37,185 (l)
![](https://rs.olm.vn/images/avt/0.png?1311)
Phương trình hóa học CaCO3 → CaO + CO2.
a) nCaO = = 0,2 mol.
Theo PTHH thì nCaCO3 = nCaO = 0,2 (mol)
b) nCaO = = 0,125 (mol)
Theo PTHH thì nCaCO3 = nCaO = 0,125 (mol)
mCaCO3 = M.n = 100.0,125 = 12,5 (g)
c) Theo PTHH thì nCO2 = nCaCO3 = 3,5 (mol)
VCO2 = 22,4.n = 22,4.3,5 = 78,4 (lít)
d) nCO2 = = 0,6 (mol)
Theo PTHH nCaO = nCaCO3 = nCO2 = 0,6 (mol)
mCaCO3 = n.M = 0,6.100 = 60 (g)
mCaO = n.M = 0,6.56 = 33,6 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2}=\dfrac{2,479}{22,4}=\dfrac{2479}{22400}mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo pt ta có: \(n_{Zn}=n_{H_2}=\dfrac{2479}{22400}mol\)\(\approx0,11mol\)
\(\Rightarrow m_{Zn}\approx7,2g\)
\(n_{HCl}=2n_{H_2}=0,22mol\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,22}{0,1}=2,2M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Ta có: \(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)
a, \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Al}=0,1.27=2,7\left(g\right)\)
b, \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=0,05\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)
c, \(n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow V_{H_2SO_4}=\dfrac{0,15}{1,5}=0,1\left(l\right)=100\left(ml\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\a, \Rightarrow n_{CaCl_2}=n_{CaCO_3}=n_{CO_2}=0,1\left(mol\right)\\ \Rightarrow a=m_{CaCO_3}=100.0,1=10\left(g\right)\\b,n_{HCl}=2.n_{CO_2}=2.0,1=0,2\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,2}{2}=0,1\left(l\right)\\ c,m_{CaCl_2}=111.0,1=11,1\left(g\right)\)
\(n_{CO2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Pt : \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O|\)
1 2 1 1 1
0,1 0,2 0,1 0,1
a) \(n_{CaCO3}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
⇒ \(m_{CaCO3}=0,1.100=10\left(g\right)\)
b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
\(V_{HCl}=\dfrac{0,2}{2}=0,1\left(l\right)\)
c) \(n_{CaCl2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
⇒ \(m_{CaCl2}=0,1.111=11,1\left(g\right)\)
Chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
Kudo làm sai , anh làm lại nha!
\(a,n_{BaCO_3}=\dfrac{49,25}{197}=0,25\left(mol\right)\\ PTHH:BaCO_3\rightarrow\left(t^o\right)BaO+CO_2\\ n_{CO_2\left(LT\right)}=n_{BaO\left(LT\right)}=n_{BaCO_3}=0,25\left(mol\right)\\ \Rightarrow n_{BaO\left(TT\right)}=n_{CO_2\left(TT\right)}=0,6.0,25=0,15\left(mol\right)\\ V_{CO_2\left(TT,đktc\right)}=0,15.22,4=3,36\left(l\right)\\ b,BaCO_3\rightarrow\left(t^o\right)BaO+CO_2\\ n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ \Rightarrow n_{BaCO_3\left(LT\right)}=n_{CO_2}=0,15\left(mol\right)\\ \Rightarrow n_{BaCO_3\left(TT\right)}=\dfrac{0,15}{80\%}=0,1875\left(mol\right)\\ m_{BaCO_3\left(TT\right)}=a=0,1875.197=\dfrac{591}{16}\left(g\right)\\ c,n_{BaCO_3}=\dfrac{49,25}{197}=0,25\left(mol\right)\)
\(Đặt:n_{BaCO_3\left(tg\right)}=j\left(mol\right)\left(a>0\right) \\Mà:m_{rắn}=42,65\\ \Leftrightarrow137j+\left(49,25-197j\right)=42,65\\ \Leftrightarrow j=11\%\\ \Rightarrow H=\dfrac{j}{0,25}.100\%=\dfrac{0,11}{0,25}.100=44\%\)
`#3107.101107`
n của \(\text{CO}_2\):
\(\text{n}_{\text{CO}_2}=\dfrac{\text{V}}{24,79}=\dfrac{2,9748}{24,79}=0,12\left(\text{mol}\right)\)
PTHH: \(\text{C}+\text{O}_2\rightarrow\text{CO}_2\)
`\rightarrow`\(\text{n}_{\text{C}}=\text{n}_{\text{CO}_2}=0,12\text{ mol}\)
- Chỉ có `80%` khối lượng a tham gia vào phản ứng
m của C:
\(a=\text{n}_{\text{C}}\cdot\text{M}_{\text{C}}=\dfrac{0,12\cdot12}{80\%}=1,8\left(\text{g}\right).\)