Câu 4: 4,4g CO2 ở 250C, P = 1bar có thể tích bao nhiêu?
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a, \(\left\{{}\begin{matrix}\%V_{SO_2}=\dfrac{3}{3+2}.100\%=60\%\\\%V_{O_2}=40\%\end{matrix}\right.\)
\(\dfrac{V_{SO_2}}{V_{O_2}}=\dfrac{3}{2}=\dfrac{n_{SO_2}}{n_{O_2}}\) ⇒ nSO2 = 3x → nO2 = 2x
\(\Rightarrow\left\{{}\begin{matrix}\%m_{SO_2}=\dfrac{3x.64}{3x.64+2x.32}.100\%=75\%\\\%m_{O_2}=25\%\end{matrix}\right.\)
b, \(2SO_2+O_2\underrightarrow{^{t^o,V_2O_5}}2SO_3\)
Xét tỉ lệ: \(\dfrac{3x}{2}< \dfrac{2x}{1}\), ta được O2 dư.
Pư: a____1/2a_____a (mol)
Sau: 3x-a___2x-1/2a___a (mol)
\(\Rightarrow\dfrac{\left(3x-a\right).64+\left(2x-\dfrac{1}{2}a\right).32+80a}{3x-a+2x-\dfrac{1}{2}a+a}=32.2\)
⇒ a = 2x
\(\Rightarrow H=\dfrac{2x}{3x}.100\%\approx66,67\%\)
a, K2SO4: potassium sulfate
NH4Cl: ammonium chloride
Ca(H2PO4)2: calcium dihydrogen phosphate
KNO3: potassium nitrate
K3PO4: potassium phosphate
NaCl: sodium chloride
FeSO4: iron (II) sulfate
Gọi: \(\left\{{}\begin{matrix}n_{N_2O_5}=x\left(mol\right)\\n_{CO_2}=y\left(mol\right)\end{matrix}\right.\)
⇒ 108x + 44y = 27,2 (1)
Mà: \(d_{A/H_2}=34\Rightarrow M_A=34.2=68\left(g/mol\right)\)
\(\Rightarrow n_A=\dfrac{27,2}{68}=0,4\left(mol\right)=x+y\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,15\left(mol\right)\\y=0,25\left(mol\right)\end{matrix}\right.\)
BTNT O, có: nO (trong A) = 5nN2O5 + 2nCO2 = 1,25 (mol)
Ta có: nO (trong Fe2(SO4)3) = 12nFe2(SO4)3
⇒ 12nFe2(SO4)3 = 1,25
⇒ nFe2(SO4)3 = 5/48 (mol)
⇒ mFe2(SO4)3 = 5/48.400 = 125/3 (g)
Gọi: nMg = x (mol), nZn = 2x (mol)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo PT: \(n_{H_2}=n_{Mg}+n_{Zn}=x+2x=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
⇒ x = 0,1 (mol)
⇒ nMg = 0,1 (mol), nZn = 0,2 (mol)
⇒ mhh = 0,1.24 + 0,2.65 = 15,4 (g)
Theo PT: \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)
\(\Rightarrow V_{HCl}=\dfrac{0,6}{2}=0,3\left(l\right)\)
\(n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,1}{0,3}=\dfrac{1}{3}\left(M\right)\)
\(n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow C_{M_{ZnCl_2}}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)
\(n_{CO_2}=\dfrac{4,4}{44}=0,1\left(mol\right)\)
\(V_{CO_2}=0,1\cdot24,79=2,479\left(l\right)\)