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Áp dụng công thức: m = M.n
=> Khối lượng của 0,2 mol Al là:
mAl = 0,2 x 27 = 5,4 gam
Câu 24 : B
n O = 2n CO2 = 0,1.2 = 0,2(mol)
Câu 25 : D
n SO2 = 6,4/64 = 0,1(mol)
Câu 26 : C
n SO2 = 64/64 = 1(mol)
V SO2 = 1.22,4 = 22,4(lít)
Câu 27 : E
Câu 28 : D
Câu 29 A
Câu 30 : C
Câu 31: B
Câu 32 : C
Câu 33 : C
4.
a) \(V_{SO_2}=0.5\cdot22.4=11.2\left(l\right)\)
b) \(V_{CH_4}=\dfrac{3.2}{16}\cdot22.4=4.48\left(l\right)\)
c) \(V_{N_2}=\dfrac{0.9\cdot10^{23}}{6\cdot10^{23}}\cdot22.4=3.36\left(l\right)\)
5.
a) \(m_{Al}=0.1\cdot27=2.7\left(g\right)\)
b) \(m_{Cu\left(NO_3\right)_2}=0.3\cdot188=56.4\left(g\right)\)
c) \(m_{Na_2CO_3}=\dfrac{1.2\cdot10^{23}}{6\cdot10^{23}}\cdot106=21.2\left(g\right)\)
d) \(m_{CO_2}=\dfrac{8.96}{22.4}\cdot44=17.6\left(g\right)\)
e) \(m_K=0.5\cdot2\cdot39=39\left(g\right)\\ m_C=0.5\cdot12=6\left(g\right)\\ m_O=0.5\cdot3\cdot16=24\left(g\right)\)
N phân tử = 1 mol phân tử
\(\Rightarrow n_{O2}=1mol;n_{N_2}=2mol;n_{CO_2}=1,5mol\)
\(\Rightarrow m_{hh}=1.32+2.28+1,5.44=154g\)
b. \(m_{hh}=0,1.56+0,2.64+0,3.65+0,25.27=44,65g\)
c. \(n_{O_2}=\dfrac{2,24}{22,4}=0,1mol\)
\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\)
\(n_{HCl}=\dfrac{6,72}{22,4}=0,3mol\)
\(n_{CO_2}=\dfrac{0,56}{22,4}=0,025mol\)
\(\Rightarrow m_{hh}=0,1.32+0,05.2+0,3.36,5+0,025.44=15,35g\)
a) \(m_X=0,1\cdot64+0,2\cdot27+0,3\cdot24=19\left(g\right)\)
\(\Rightarrow\%m_{Cu}=\dfrac{0,1\cdot64\cdot100}{19}=34\%\)
\(\Rightarrow\%m_{Al}=\dfrac{0,2\cdot27\cdot100}{19}=28\%\)
\(\Rightarrow\%m_{Mg}=100\%-34\%-28\%=38\%\)
b) \(Cu+2HCl\rightarrow CuCl_2+H_2\) (1)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\) (2)
\(Mg+2HCl\rightarrow MgCl_2+H_2\) (3)
\(n_{HCl\left(1\right)}=2n_{Cu}=2\cdot0,1=0,2\left(mol\right)\)
\(n_{HCl\left(2\right)}=\dfrac{6\cdot n_{Al}}{2}=3\cdot0,2=0,6\left(mol\right)\)
\(n_{HCl\left(3\right)}=2n_{Mg}=2\cdot0,3=0,6\left(mol\right)\)
\(\Rightarrow m_{HCl}=\left(0,2+0,6+0,6\right)\cdot36,5=51,1\left(g\right)\)
\(n_{H_2\left(1\right)}=n_{Cu}=0,1\left(mol\right);n_{H_2\left(2\right)}=\dfrac{3\cdot0,2}{2}=0,3\left(mol\right);n_{H_2\left(3\right)}=n_{Mg}=0,3\left(mol\right)\)
\(V_{H_2\left(dkc\right)}=\left(0,1+0,3+0,3\right)\cdot24,79=17,353\left(l\right)\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH :
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
0,1 0,1
-> n H2SO4 = 0,1 mol
---> B
a, \(\overline{M}=\dfrac{0,1.44+0,2.28}{0,1+0,2}\approx33,33\left(g/mol\right)\)
b, \(\overline{M}=\dfrac{0,2.28+0,3.2}{0,2+0,3}=12,4\left(g/mol\right)\)
c, \(\overline{M}=\dfrac{0,1.28+0,2.30+0,2.44}{0,1+0,2+0,2}=35,2\left(g/mol\right)\)
d, \(\overline{M}=\dfrac{0,2.56+0,1.24+0,1.27}{0,2+0,1+0,1}=40,75\left(g/mol\right)\)
a) 1,5 x 6.1023 = 9.1023 hay 1,5N (nguyên tử Al).
b) 0,5 x 6.1023 = 3.1023 hay 0,5N (phân tử H2).
c) 0,25 x 6.1023 = 1,5.1023 hay 0,25N (phân tử NaCl).
d) 0,05 x 6.1023 = 0,3.1023 hay 0,05N (phân tử H2O).
mAl=nAl.MAl=1,5.27=40,5(g)
mH2=nH2.MH2=0,5.2=1(g)
mNaCl=nNaCl.MNaCl=0,25.(23+35,5)=14,625(g)
mH2O=nH2O.MH2O=0,05.18=0,9
C
C