Mg (s) + 2HCl (aq) → H2 (g) + MgCl2 (aq)
A:
Moles Mg: 0.050
Moles HCl: 0.050

Mass of Hydrogen gas and the limiting reactant

Mass of hydrogen gas formed from the given reaction is 0.05 grams and the limiting reactant is HCl.

Explanation:

Data given:

moles of Mg = 0.050 moles

moles of HCl = 0.050 moles

mass of hydrogen gas formed =?

atomic mass of hydrogen gas = 2 grams/mole

limiting reactant = ?

Balanced chemical equation is given by:

Mg (s) + 2HCl (aq) → H2 (g) + Mg (aq)

from the reaction let us deduce the limiting reagent as

1 mole of Mg gave 1 mole of hydrogen gas

0.050 moles will give 0.050 moles of hydrogen gas

mass of hydrogen gas  = atomic mass x number of moles

mass of hydrogen gas = 2 x 0.050

                                       = 0.1 grams of hydrogen gas produced.

Now from 2 moles of HCl 1 mole of hydrogen gas is formed

so, 0.050 moles of HCl will give x moles of hydrogen gas

=

2x = 0.050

x = 0.025 moles

mass of hydrogen gas  = atomic mass x number of moles

mass of hydrogen gas  = 0.025 x 2

                                      = 0.05 grams

mass = atomic mass x number of moles

The limiting reactant is HCl as it produced less amount of product after getting fully consumed in the reaction.

the product formed depends on the limiting reagent and the value is 0.05 grams.