Ta có: \(m_{H_2SO_4}=160.98\%=156,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{156,8}{98}=1,6\left(mol\right)\)
\(n_S=\dfrac{9,6}{32}=0,3\left(mol\right)\)
\(4H_2SO_4+6e\rightarrow3SO_4^{2-}+S+4H_2O\)
1,2_______________0,9___0,3 (mol)
\(2H_2SO_4+2e\rightarrow SO_4^{2-}+SO_2+2H_2O\)
0,4______________0,2_____0,2 (mol)
⇒ m muối = mA + mSO42- = 26,92 + (0,9 + 0,2).96 = 132,52 (g)
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