Câu 2

PTK(CaCO3)=100 đvC

PTK ( N2)=28đvC

PTK (4KCl)=191,5 đvC

PTK (2BaSO4)= 370đvC

“Phân tử khối bằng tổng khối lượng của các nguyên tửu trong phân tử”

A l 2 O 3  (M = 27.2 + 16.3 = 102 đvC )

A l 2 ( S O ₄ ) 3  (M = 342 đvC ) F e ( N O 3 ) 3  ( M = 242 đvC )

N a 3 P O 4  (M = 164 đvC ) C a ( H 2 P O 4 ) 2    ( M = 234 đvC )

B a 3 ( P O 4 ) 2    (M = 601 đvC ) Z n S O 4  ( M = 161 đvC )

AgCl (M = 143,5 đvC ) NaBr ( M = 103 đvC )

PTK SO₃=32+48=80 đvC

PTK NaHCO₃=23+1+12+48=84 đvC

PTK Al₂(SO₄)₃=54+96+192=342 đvC

PTK PbSO₄=207+32+64=303 đvC

PTK CH₄=12+4=16 đvC

\(a,n_{\left(NH_4\right)_3PO_4}=0,6\left(mol\right)\\ \Rightarrow n_N=0,6.3=1,8\left(mol\right)\Rightarrow m_N=1,8.14=25,2\left(g\right)\\ n_H=4.3.0,6=7,2\left(mol\right)\Rightarrow m_H=7,2.1=7,2\left(g\right)\\ n_P=n_{hc}=0,6\left(mol\right)\Rightarrow m_P=0,6.31=18,6\left(g\right)\\ n_O=4.0,6=2,4\left(mol\right)\Rightarrow m_O=2,4.16=38,4\left(g\right)\)

\(b,n_S=\dfrac{6,4}{32}=0,2\left(mol\right)\Rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.0,2=\dfrac{1}{15}\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=342.\dfrac{1}{15}=22,8\left(g\right)\\ c,n_{Al_2\left(SO_4\right)_3}=\dfrac{20,52}{342}=0,06\left(mol\right)\\ n_O=4.3.0,06=0,72\left(mol\right)\\ \Rightarrow n_{CO_2}=\dfrac{0,72}{2}=0,36\left(mol\right)\Rightarrow V_{CO_2\left(đktc\right)}=0,36.22,4=8,064\left(l\right)\)

\(M_{5K}=5\cdot39=195\left(đvc\right)\)

\(M_{6N_2}=6\cdot28=168\left(đvc\right)\)

\(M_{3O_2}=3\cdot32=96\left(đvc\right)\)

\(M_{2Al_2\left(SO_4\right)_3}=2\cdot342=684\left(đvc\right)\)

PTK: + H₃PO₄ = 191 đvC

         + Na₂CO₃ = 106 đvC

         + SO₃ = 80 đvC

         + NaOH = 49 đvC

         + Al₂(SO₄)₃ = 342 đvC

PTK: + H₃PO₄ = 191 đvC

         + Na₂CO₃ = 106 đvC

         + SO₃ = 80 đvC

         + NaOH = 49 đvC

         + Al₂(SO₄)₃ = 342 đvC

a, \(n_{Al_2\left(SO_4\right)_3}=\dfrac{42,75}{342}=0,125\left(mol\right)\)

\(n_O=12n_{Al_2\left(SO_4\right)_3}=1,5\left(mol\right)\)

b, \(n_{O\left(Al_2O_3\right)}=3n_{Al_2O_3}=2n_{O\left(Al_2\left(SO_4\right)_3\right)}=3\left(mol\right)\)

\(\Rightarrow n_{Al_2O_3}=1\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=102\left(g\right)\)

PTK: 27.2 + ( 32.3 + 16.12 ) = 310 đvC

PTK=27.2+32.3+16.12=342 dvC

\(PTK_{Ca\left(OH\right)_2}=NTK_{Ca}+2.\left[NTK_O+NTK_H\right]=40+2.\left(16+1\right)=74\left(đ.v.C\right)\\ PTK_{Fe\left(OH\right)_3}=NTK_{Fe}+3.\left[NTK_O+NTK_H\right]=56+3.\left(16+1\right)=107\left(đ.v.C\right)\\ PTK_{KNO_3}=NTK_K+NTK_N+3.NTK_O=39+14+3.16=101\left(đ.v.C\right)\\ PTK_{Fe_2O_3}=2.NTK_{Fe}+3.NTK_O=2.56+3.16=160\left(đ.v.C\right)\)

\(PTK_{N_2O_5}=2.NTK_N+5.NTK_O=2.14+5.16=108\left(đ.v.C\right)\\ PTK_{MgSO_4}=NTK_{Mg}+NTK_S+4.NTK_O=24+32+4.16=120\left(đ.v.C\right)\\ PTK_{Al_2\left(SO_4\right)_3}=2.NTK_{Al}+3.\left[NTK_S+3.4.NTK_O\right]\\ =2.27+3.\left(32+3.4.16\right)=342\left(đ.v.C\right)\\ PTK_{BaCO_3}=NTK_{Ba}+NTK_C+3.NTK_O=137+12+3.16=197\left(đ.v.C\right)\)