Question

Answer 1

Mg+2HCl->Mgcl2+H2

0,2--------------0,2-----0,2 mol

n Mg=4,8\24=0,2 mol

=>m MgCl2=0,2.95=19g

=>VH2=0,2.22,4=4,48l

Answer 2

a,\(n_{HCl}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

Mol:     0,2                   0,2           0,2

b,\(m_{MgCl_2}=0,2.95=19\left(g\right)\)

c,\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

Answer 3

a. PT: Mg + 2HCl ---> MgCl₂ + H₂

b. Ta có: n_Mg = \(\dfrac{4,8}{24}=0,2\left(mol\right)\)

Theo PT: \(n_{MgCl_2}=2.n_{Mg}=0,4\left(mol\right)\)

=> \(m_{MgCl_2}=0,4.95=38\left(g\right)\)

c. Theo PT: \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)

=> \(V_{H_2}=0,2.22,4=4,48\left(lít\right)\)