a) Fe + H2SO4 → FeSO4 + H2
b) \(n_{H_2}=\dfrac{13,48}{22,4}=\dfrac{337}{560}\left(mol\right)\)
Theo PT: \(n_{Fe}pư=n_{H_2}=\dfrac{337}{560}\left(mol\right)\)
\(\Rightarrow m_{Fe}pư=\dfrac{337}{560}\times56=33,5\left(g\right)\)
c) Theo PT: \(n_{H_2SO_4}pư=n_{H_2}=\dfrac{337}{560}\left(mol\right)\)
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{337}{560}\div0,2\approx3\left(M\right)\)
nH2 = \(\dfrac{13,48}{22,4}\) \(\approx\)0,6 mol
Fe + H2SO4 -> FeSO4 + H2
0,6<-0,6<----------------0,6
=>mFe = 0,6 . 56 = 33,6 g
V = 200 ml = 0,2 (l) => CM(H2SO4) = \(\dfrac{0,6}{0,2}\) = 3M
