Question

Giúp mình

Answer 1

\(n_{Mg}=\dfrac{5,28}{24}=0,22\left(mol\right)\)

a) \(Mg+2HCl\xrightarrow[]{}MgCl_2+H_2\uparrow\)

      0,22 → 0,44   →  0,22 → 0,22

\(V_{H_2}=0,22\cdot22,4=4,928\left(l\right)\)

b) \(m_{HCl}=0,44\cdot36,5=16,06\left(g\right)\)

\(C\%_{HCl}=\dfrac{16,06}{200}\cdot100\%=8,03\%\)

c) \(m_{H_2O}=200-16,06=183,94\left(g\right)\)

\(m_{MgCl_2}=0,22\cdot95=20,09\left(g\right)\)

\(m_{dd\text{ sau}}=183,94+20,09=204,03\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{20,09}{204,03}\cdot100\approx9,85\%\)