trả lời giúp mik ik ạ.3

\(M\) _{\(C_2H_6\)}\(=12.2+1.6=30\) \(đvC\)

\(M\) _{\(MgCO_3\)} \(=24+12+16.3=84\) \(đvC\)

\(M\) _{\(Al_2\left(SO_4\right)_3\)} \(=27.2+32.3+16.12=342\) \(đvC\)

C2H6 là 12*2+1*6=24+6=30 đvC

MgCO3 là 24+12+3*16=84 đvC

Al2(SO4)3 là 27*2+32*3+16*12=242 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

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

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

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

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

\(M_{Cl_2}=35.5\cdot2=71\left(đvc\right)\)

\(M_{H_2SO_4}=2\cdot1+32+64=98\left(đvc\right)\)

\(M_{Cu\left(NO_3\right)_2}=64+62\cdot2=188\left(đvc\right)\)

\(M_{Al_2\left(SO_4\right)_3}=27\cdot2+\left(32+16\cdot4\right)\cdot3=342\left(đvc\right)\)

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

\(PTK_{H_3PO_4}=1.3+31+16.4=98\left(đvC\right)\)

\(PTK_{Na_2CO_3}=23.2+12+16.3=106\left(đvC\right)\)

\(PTK_{SO_3}=32+16.3=80\left(đvC\right)\)

\(PTK_{NaOH}=23+16+1=40\left(đvC\right)\)

\(PTK_{Al_2\left(SO_4\right)_3}=27.2+\left(32+16.4\right).3=342\left(đvC\right)\)

D.

D

D

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

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

a) biết \(PTK_{Al_2\left(SO_4\right)_3}=2.27+\left(1.32+4.16\right).3=342\left(đvC\right)\)

vậy \(PTK_C=342-178=164\left(đvC\right)\)

b) ta có:

\(3X+PO_4=164\)

\(3X+1.31+4.16=164\)

\(3X+31+64=164\)

\(3X+95=164\)

\(3X=164-95=69\)

\(X=\dfrac{69}{3}=23\left(đvC\right)\)

\(\Rightarrow X\) là \(Na\left(Natri\right)\)