\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\)
a) \(Mg+H_2SO_4\text{ (loãng)}\xrightarrow[]{}MgSO_4+H_2\uparrow\)
0,25 → 0,25
\(V_{H_2}=0,25\cdot22,4=5,6\left(l\right)\)
b) \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\)
\(Fe_2O_3+3H_2\xrightarrow[]{t^\circ}2Fe+3H_2O\uparrow\)
\(\dfrac{1}{12}\) ← 0,25 → \(\dfrac{1}{6}\)
\(m_{Fe}=\dfrac{1}{6}\cdot56=\dfrac{28}{3}\left(g\right)\)
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