\(n_{Cl_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ pthh:2R+Cl_2\rightarrow2RCl\) 
          0,1     0,05     0,1 
\(M_R=\dfrac{2,3}{0,1}=23\left(\dfrac{g}{mol}\right)\) 
mà R hóa trị I 
-> R là Na 
\(pthh:2Na+Cl_2\rightarrow2NaCl\) 
           0,1       0,05     0,1 
\(m_{NaCl}=58,5.0,1=5,85\left(g\right)\)

\(a,SO_3;ptk=32+16.3=80\left(\text{đ}vC\right) -g,P_2O_5;ptk=31.2+16.5=142\left(\text{đ}vC\right)\\ b,K_2SO_4; ptk=39.2+32+16.4=174\left(\text{đ}vC\right)-h,Zn\left(OH\right)_2;ptk:65+\left(16+1\right).2=99\left(\text{đ}vC\right)\\ c,Ba_3\left(PO_4\right)_2; ptk=137.3+\left(31+16.4\right).2=601\left(\text{đ}vC\right)-h,MgSO_4;ptk=24+32+16.4=120\left(\text{đ}vC\right)\\ e,Al\left(NO_3\right)_3;ptk:27+\left(14+16.3\right).3=213\left(\text{đ}vC\right)-f,NaOH;ptk=23+16+1=40\left(\text{đ}vC\right)\)

Tìm giá trị nguyên của x để B có giá trị nguyên 
\(B=\dfrac{10x^2-7x-5}{2x-3}\)

(5+x) x7  = 81-4 
(5+x) x7 = 77 
(5+x) = 11 
x= 6 

\(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\\ pthh:Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\) 
          0,2            1,2 
\(m_{HCl}=1,2.36,5=43,8\left(g\right)\\ m_{\text{dd}HCl}=\dfrac{43,8.100}{7,3}=600\left(g\right)\\ V_{\text{dd}\left(HCl\right)}=\dfrac{600}{1,18}=508,474\left(ml\right)\)