đốt cháy hoàn toàn 6,8g một hợp chất vô cơ A chỉ thu được 4,48 lít SO2 và 3,6g H2O. tìm cthh của A
\(n_{SO_2} = \dfrac{4,48}{22,4} = 0,2(mol)\\ n_{H_2O} = \dfrac{3,6}{18} = 0,2(mol) \)
BTKL :
\(m_{O_2} = m_{SO_2} + m_{H_2O} - m_A = 0,2.64 + 3,6-6,8=9,6(gam) \\ \Rightarrow n_{O_2} = 0,3\ mol\)
BTNT với O :
\(n_O = 2n_{SO_2} + n_{H_2O} - 2n_{O_2} = 0,2.2 + 0,2 - 0,3.2 = 0\).Chứng tỏ A không chứa oxi.
\(n_S = n_{SO_2} = 0,2\ mol\\ n_H = 2n_{H_2O} = 0,4\ mol\)
Ta có :
\(n_S : n_H = 0,2 : 0,4 = 1 : 2\)
Vậy CTHH của A : H2S
PTHH: \(3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\)
Ta có: \(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)
\(\Rightarrow n_{FeCl_3}=n_{Fe\left(OH\right)_3}=\dfrac{1}{12}\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}m_{FeCl_3}=\dfrac{1}{12}\cdot162,5\approx13,54\left(g\right)\\m_{Fe\left(OH\right)_3}=\dfrac{1}{12}\cdot107\approx8,92\left(g\right)\end{matrix}\right.\)
nNaOH = m/M = 10/(23 +16 + 1) = 0,25 (mol)
Ta có PTHH: 3NaOH + FeCl3 ------> Fe(OH)3 + 3NaCl
Theo PT: 3 - 1 - 1 (mol)
BC: 0.25 - 0.083 - 0.083 (mol)
Suy ra: mFeCl3 = n x M = 0.083 x (56 + 35,5 x 3) = 13,4875 (g)
mFe(OH)3 = n x M = 0,083 x (56+17 x 3) = 8,881 (g)
PTHH : 3NaOH + FeCl3 ----> Fe(OH)3 + 3NaCl
a) \(n_{\text{NaOH}}\) = \(\dfrac{m}{M}\) = 0.25 (mol)
Có : \(n_{\text{NaOH}}\) = \(\dfrac{1}{3}\) \(n_{FeCl_3}\)
--> \(n_{FeCl_3}\) = \(\dfrac{1}{12}\) (mol)
=> \(m_{FeCl_3}\) = n . M \(\approx\) 13.5 (g)
b) Ta có : \(n_{FeCl_3}\) = \(n_{Fe\left(OH\right)_3}\)
--> \(n_{Fe\left(OH\right)_3}\) = \(\dfrac{1}{12}\) (mol)
=> \(m_{Fe\left(OH\right)_3}\) = n . M \(\approx\) 9 (g)
PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
a) Ta có: \(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(\Rightarrow n_{Fe}=0,45mol\) \(\Rightarrow m_{Fe}=0,45\cdot56=25,2\left(g\right)\)
b)
+) Cách 1:
Theo PTHH: \(n_{Fe_3O_4}=\dfrac{1}{2}n_{O_2}=0,15mol\)
\(\Rightarrow m_{Fe_3O_4}=0,15\cdot232=34,8\left(g\right)\)
+) Cách 2:
Ta có: \(m_{O_2}=0,3\cdot32=9,6\left(g\right)\)
Bảo toàn khối lượng: \(m_{Fe_3O_4}=m_{O_2}+m_{Fe}=25,2+9,6=34,8\left(g\right)\)
n O2=6,72/22,4=0,3(mol)
3Fe+2O2-------->Fe3O4
TPT:3. 2. 1
TB:. ? 0,3. ?
Theo phương trình và bài ra ta có:
n Fe=0,3×3/2=0,45(mol)
m Fe=0,45×56=25,2(g)
b). C1: Ta có :
n Fe3O4=0,3×1/2=0,15 (mol)
m Fe3O4=0,15×232=34,8(g)
C2:. Ta có :
m O2 = 0,3×32=9,6(mol)
m Fe + m O2 =m Fe3O4
25,2+9,6=34,8(g)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
a) Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(\Rightarrow n_{HCl}=0,6mol\) \(\Rightarrow m_{HCl}=0,6\cdot36,5=21,9\left(g\right)\)
b) Theo PTHH: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3mol\)
\(\Rightarrow V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\)
c)
+) Cách 1:
Theo PTHH: \(n_{AlCl_3}=n_{Al}=0,2mol\) \(\Rightarrow m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\)
+) Cách 2:
Ta có: \(m_{H_2}=0,3\cdot2=0,6\left(g\right)\)
Bảo toàn khối lượng: \(m_{AlCl_3}=m_{Al}+m_{HCl}-m_{H_2}=26,7\left(g\right)\)
\(n_{Al_2O_3}=\dfrac{10.2}{102}=0.1\left(mol\right)\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.1..........0.6\)
\(V_{ddHCl}=\dfrac{0.6}{2}=0.3\left(l\right)\)
a)
\(n_{Al} = \dfrac{2,7}{27} = 0,1(mol)\\ n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)\)
\(4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\)
Ta thấy :
\(\dfrac{n_{Al}}{4} = 0,025 < \dfrac{n_{O_2}}{3} = 0,1\) nên O2 dư
Theo PTHH :
\(n_{Al_2O_3} = 0,5n_{Al} = 0,05(mol)\\ n_{O_2\ pư} = \dfrac{3}{4}n_{Al} = 0,075(mol)\)
Suy ra :
\(m_{Al_2O_3} = 0,05.102 = 5,1(gam)\\ m_{O_2\ dư} = (0,3 - 0,075).32 = 7,2(gam)\)
PTHH: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
a) Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{4}< \dfrac{0,3}{3}\) \(\Rightarrow\) Oxi còn dư, Al phản ứng hết
\(\Rightarrow\left\{{}\begin{matrix}n_{Al_2O_3}=0,05mol\\n_{O_2\left(dư\right)}=0,225mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=0,225\cdot32=7,2\left(g\right)\\m_{Al_2O_3}=0,05\cdot102=5,1\left(g\right)\end{matrix}\right.\)
b) Theo PTHH: \(n_{O_2\left(pư\right)}=0,075mol\)
\(\Rightarrow V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\)
Vì Oxi chiếm khoảng 20% thể tích không khí
\(\Rightarrow V_{kk}=\dfrac{1,68}{20\%}=8,4\left(l\right)\)
\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)
\(n_{O_2}=0.3\left(mol\right)\)
\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)
\(0.1.....0.075.....0.05\)
\(m=m_{Al_2O_3}+m_{O_2\left(dư\right)}=0.05\cdot102+\left(0.3-0.075\right)\cdot32=12.3\left(g\right)\)
\(V_{kk}=5V_{O_2}=5\cdot0.075\cdot22.4=8.4\left(l\right)\)
Bài 1 :
a)
\(2Cu + O_2 \xrightarrow{t^o} 2CuO\)
b)
Ta có :
\(n_{Cu} = \dfrac{32}{64} = 0,5(mol)\)
Theo PTHH : \(n_{O_2} = 0,5n_{Cu} = 0,25(mol)\\ \Rightarrow V_{O_2} = 0,25.22,4 = 5,6(lít)\)
c) Ta có : \(n_{CuO} = n_{Cu} = 0,5(mol)\Rightarrow m_{CuO} = 0,5.80 = 40(gam)\)
Bài 2 :
\(Zn + H_2SO_4 \to ZnSO_4 + H_2\)
Theo PTHH :
\(n_{H_2SO_4} = n_{H_2} = n_{Zn} =\dfrac{13}{65} = 0,2(mol)\)
Suy ra :
\(V_{H_2} = 0,2.22,4 = 4,48(lít)\\ m_{H_2SO_4} = 0,2.98 = 19,6(gam)\)
nNa2CO3 = 0,1 mol
Na2CO3 + CaCl2 → CaCO3 + 2NaCl
0,1..............0,1................0,1............0,2
⇒ mCaCO3 = 0,1.100 = 10 (g)
⇒ mNaCl = 0,2.58,5 = 11,7 (g)
8tk
a)
\(n_{H_2} = \dfrac{1,12}{22,4} = 0,05(mol)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\)
Ta thấy :
\(n_{HCl} = 0,2 > 2n_{H_2} = 0,1\) nên HCl dư.
Theo PTHH : \(n_{Al} = \dfrac{2}{3}n_{H_2} = \dfrac{1}{30}(mol)\\ \Rightarrow m_{Al} = \dfrac{1}{30}.27 = 0,9(gam)\)
Ta có :
\(n_{HCl\ pư} = 2n_{H_2} = 0,1(mol)\\ \Rightarrow n_{HCl\ dư} = 0,2 - 0,1 = 0,1(mol)\\ \Rightarrow m_{HCl\ dư} = 0,1.36,5 = 3,65(gam)\)