Question

How many grams of magnesium are needed to prepare 4.60 L of H2 at 760 mmHg and 28 ∘C?

Answer 1

$Mg+ 2HCl\to MgCl_2+H_2$

Moles of Hydrogen gas obtained =$\frac{4.6}{22.4}=0.205\ mole$

Moles of Mg required = 0.205 moles

Mass of Mg required = $0.205\times 24.3= 5\ gm$