biết \(M_{H_2}=1.2=2\left(đvC\right)\)
vậy \(M_A=2.87=174\left(đvC\right)\)
ta có:
\(2K+1S+4X=174\)
\(2.39+1.32+4X=174\)
\(78+32+4X=174\)
\(110+4X=174\)
\(4X=174-110\)
\(4X=64\)
\(X=\dfrac{64}{4}=16\left(đvC\right)\)
\(\rightarrow X\) là \(O\left(Oxi\right)\)
\(\rightarrow CTHH:K_2SO_4\)
