Question

Answer 1

6.3/

a/ \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

PTHH: Mg + H₂SO₄ → MgSO₄ + H₂

Mol:    0,15      0,15         0,15          0,15

\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

b/ \(m_{MgSO_4}=0,15.120=18\left(g\right)\)

c/ \(m_{ddH_2SO_4}=\dfrac{0,15.98.100}{9,8}=150\left(g\right)\)

Answer 2

6.4/

a/ \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: Fe + H₂SO₄ → FeSO₄ + H₂

Mol:     0,15                    0,15         0,15

\(m_{Fe}=0,15.56=8,4\left(g\right)\)

b/ \(m_{FeSO_4}=0,15.152=22,8\left(g\right)\)

c/ \(C_{M_{ddFeSO_4}}=\dfrac{0,15}{0,5}=0,3M\)

Answer 3