How many moles of H 2 can be formed if a 3.24 g sample of M g reacts with excess H C l ?

0.135 mole of H2 can be formed.

The reaction of Mg with HCl is:

→

The moles of H₂ formed will be:

=

=

=

Thus the above answer is right.

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0.135 mole of H2.

Explanation:

We'll begin by calculating the number of mole in 3.24 g of Mg. This can be obtained as follow:

Mass of Mg = 3.24 g

Molar mass of Mg = 24 g/mol

Mole of Mg =?

Mole = mass /Molar mass

Mole of Mg = 3.24/24

Mole of Mg = 0.135 mole

Next, we shall write the balanced equation for the reaction. This is illustrated below:

Mg + 2HCl —> MgCl2 + H2

From the balanced equation above,

1 mole of Mg reacted to produce 1 mole of H2.

Finally, we shall determine the number of mole of H2 produced by reacting 3.24 g (i.e 0.135 mole) of Mg. This can be obtained as follow:

From the balanced equation above,

1 mole of Mg reacted to produce 1 mole of H2.

Therefore, 0.135 mole of Mg will also react to produce 0.135 mole of H2.

Thus, 0.135 mole of H2 can be obtained from the reaction.

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

the answer would be a