Question

Answer 1

a/ \(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)

PTHH: MgO + 2HCl → MgCl₂ + H₂O

Mol:      0,2         0,4             0,2

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

b/ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{14,6.100}{4}=365\left(g\right)\)

Answer 2

\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\\ MgO+2HCl\rightarrow MgCl_2+H_2O\\ 0,2........0,4..........0,2........0,2\left(mol\right)\\ m_{MgCl_2}=95.0,2=19\left(g\right)\\ b.m_{ddHCl}=\dfrac{0,4.36,5.100}{4}=365\left(g\right)\)