A student placed 0.0243 g of magnesium in 50 mL of 0.200 molar HCl solution, producing H2 gas according to the following reaction what is the volume of H2 (g) produce at 300 K and 0.800 atm?

Volume of hydrogen gas = 0.1539dm3

Explanation:

Equation of the reaction

Mg (s) + 2HCl - -> MgCl2 (aq) + H2 (g)

Molar concentration of HCl = 0.2mol/dm3

Mass of magnesium = 0.0243g

Volume of HCl = 0.05dm3

No of moles of HCl = Volume * molar concentration

= 0.05 * 0.5

= 0.01mol

No of moles of Magnesium = mass/molecular weight

Molecular weight of Magnesium = 24g/mol

No of moles of magnesium = 0.0243/24

= 0.001013mol

Using stoichiometry,

Since 1 mole of Magnesium metal reacted with 2 moles of HCl acid

HCl is the limiting reagent.

Therefore, 1 mole of hydrogen gas is given off.

0.005 mole of hydrogen is given off.

Using ideal gas law, PV = nRT

Volume = (0.005 * 0.0821 * 300) / 0.8

= 0.1539dm3