Hóa học

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ BTNT\left(Al\right):n_{Al_2O_3}.2=n_{Al}\\ \Rightarrow n_{Al_2O_3}=0,05\left(mol\right)\\VìH=80\%\\ \Rightarrow m_{Al_2O_3}=\dfrac{0,05.102}{80\%}=6,375\left(tấn\right)\)

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a) \(V_{O_2}=0,35.24,79=8,6765\left(l\right)\)

b) \(n_{Al\left(OH\right)_3}=\dfrac{23,4}{78}=0,3\left(mol\right)\)

c) Số nguyên tử Na = 2,5.6.10²³ = 15.10²³

Câu 1: 

a, \(\%m_{Na}=\dfrac{46.100\%}{106}=43,39\%\)

\(\%m_C=\dfrac{12.100\%}{106}=11,32\%\)

\(\Rightarrow\%m_O=100\%-43,39\%-11,32\%=45,29\%\)

b, \(\%m_{Ba}=\dfrac{137.100\%}{208}=65,86\%\)

\(\Rightarrow\%m_{Cl}=100\%-65,86\%=34,14\%\)

Câu 2: 

a, \(V_{N_2}=\left(\dfrac{14}{14}\right).22,4=22,4l\)

b, \(V_{NH_3}=\left(\dfrac{32}{17}\right).22,4=42,112l\)

c, \(V_{Fe}=\left(\dfrac{6,5}{56}\right).22,4=2,688l\)

d, \(V_{CO_2}=\left(\dfrac{11}{44}\right).22,4=5,6l\)

\(4FeS_2+11O_2->2Fe_2O_3+8SO_2\)

==>A

4FeS₂ + 11O₂  \(\underrightarrow{t^o}\) 2Fe₂O₃ + 8SO₂

=> Đáp án D

\(D. 4FeS_2 +11 O_2 ->2 Fe_2O3 + 8SO_2 \)