Bài 18: Mol

c)

\(n_{H_2}=\dfrac{m}{M}=\dfrac{13,2}{2}=6,6\left(mol\right)\)

d)

\(n_{SO_2}=\dfrac{m}{M}=\dfrac{12,8}{32+16\cdot2}=0,2\left(mol\right)\)

đề cho khối lượng thì tính số mol của các chất khí ơ dktc là sao vậy bn?=))

Gọi \(\left\{{}\begin{matrix}n_{O_2}=a\left(mol\right)\\n_{N_2}=b\left(mol\right)\end{matrix}\right.\left(ĐK:a,b>0\right)\)

=> \(\left\{{}\begin{matrix}m_{O_2}=32a\left(g\right)\\m_{N_2}=28b\left(g\right)\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}n_{hh.khí}=n_{O_2}+n_{N_2}=a+b=0,8\\m_{hh.khí}=m_{O_2}+m_{N_2}=32a+28b=23,2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,6\left(mol\right)\end{matrix}\right.\left(TM\right)\)

=> \(\left\{{}\begin{matrix}m_{O_2}=32a=32.0,2=6,4\left(g\right)\\m_{N_2}=m_{hh.khí}-m_{O_2}=23,2-6,4=16,8\left(g\right)\end{matrix}\right.\)

Mai kiểm tra 15' rồi hy vọng mn giúp

\(N_{Fe}=0,25\cdot6\cdot10^{23}=1,5\cdot10^{23}\)

\(N_{Al}=1,5\cdot6\cdot10^{23}=9\cdot10^{23}\\ N_{H_2}=0,5\cdot6\cdot10^{23}=3\cdot10^{23}\\ N_{NaCl}=0,25\cdot6\cdot10^{23}=1,5\cdot10^{23}\\ n_{H_2O}=\dfrac{1,2\cdot10^{22}}{6\cdot10^{23}}=0,02\left(mol\right)\\ n_{Mg}=\dfrac{0,72\cdot10^{23}}{6\cdot10^{23}}=0,12\left(mol\right)\\ n_{CO_2}=\dfrac{3\cdot10^{22}}{6\cdot10^{23}}=0,05\left(mol\right)\)

\(n_{Fe_2O_3}=\dfrac{80}{56\cdot2+16\cdot3}=0,5\left(mol\right)\\ n_{CaCO_3}=\dfrac{10}{40+12+16\cdot3}=0,1\left(mol\right)\\ n_{Na_2SO_4}=\dfrac{56,8}{23\cdot2+32+16\cdot4}=0,4\left(mol\right)\\ n_{K_2SO_4}=\dfrac{43,5}{39\cdot2+32+16\cdot4}=0,25\left(mol\right)\\ n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\\ n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\)

\(m_{CuO}=0,25\cdot\left(64+16\right)=20\left(g\right)\\ m_{BaCl_2}=0,25\cdot\left(137+35,5\cdot2\right)=52\left(g\right)\\ m_{NaCl}=0,05\cdot\left(23+35,5\right)=2,925\left(g\right)\\ n_{H_2SO_4}=0,15\cdot\left(2+32+16\cdot4\right)=14,7\left(g\right)\)

a) \(n_{H_2O}=\dfrac{1,8}{18}=0,1\left(mol\right)\)

b) \(m_{NaCl}=0,2.58,5=11,7\left(g\right)\)

c) \(V_{O_2}=0,25.22,4=5,6\left(l\right)\)

d) \(n_{CO_2}=\dfrac{0,6.10^{23}}{6.10^{23}}=0,1\left(mol\right)\Rightarrow m_{CO_2}=0,1.44=4,4\left(g\right)\)

Gọi CTHH của A là $N_xH_y$

Ta có : 

$\dfrac{14x}{82,35\%} = \dfrac{y}{17,65\%} = \dfrac{17}{100\%}$

Suy ra:  x = 1 ; y = 3

Vậy CTHH của A là $NH_3$

a) \(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

b) \(n_{CO_2}=\dfrac{6,6}{44}=0,15\left(mol\right)\)