Answer:
[tex]\large \boxed{\text{1.9 mol SrCl}_{2}}[/tex]
Explanation:
Sr + 2HCl ⟶ SrCl₂ + H₂
n/mol: 3.8
The molar ratio is 1 mol SrCl₂:2 mol HCl.
[tex]\text{Moles of SrCl}_{2} = \text{3.8 mol HCl} \times \dfrac{\text{1 mol SrCl}_{2}}{\text{2 mol HCl}}= \textbf{1.9 mol SrCl}_{2}\\\\\text{The reaction produces $\large \boxed{\textbf{1.9 mol SrCl}_{2}}$}[/tex]
