Gọi số mol Fe, Zn là a, b (mol)
PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\Rightarrow n_{H_2SO_4}=1\left(mol\right)\)
=> \(m_{H_2SO_4}=1.98=98\left(g\right)\Rightarrow m_{dd.H_2SO_4}=\dfrac{98.100}{10}=980\left(g\right)\)
mdd sau pư = 56a + 65b + 980 - 1.2
= 56a + 65b + 978 (g)
Ta có: \(n_{H_2}=a+b=1\left(mol\right)\) (1)
\(C\%_{ZnSO_4}=\dfrac{161b}{56a+65b+978}.100\%=11,6022\%\) (2)
(1)(2) => a = 0,25 (mol) ; b = 0,75 (mol)
=> mFe = 0,25.56 = 14 (g); mZn = 0,75.65 = 48,75 (g)
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