Question

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Answer 1

1) We know that there are 6.023 X 10^23 number of atoms or molecules or ions in one mole of a substance

In case of Magnesium it will have 6.023 X 10^23 number of atoms in one mole

The molar mass of Mg is 24 g/mol

so the mass of one mole of Mg will be 24g

Now moles of Mg in 35 grams = 35 / 24 = 1.458 moles

so number of atoms in 35g = moles X Na

[Where Na = Avagadro's number = 6.023 X 10^23]

number of atoms of Magnesium in 35g

                  = 1.458 X 6.023 X 10^23 = 8.78  X 10^23 atoms

2) The reaction is

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

So here one mole of Mg is reacting with two moles of HCl

The molar mass of Mg is 24 g/mol

so the mass of one mole of Mg will be 24g

The moles of Mg in 25 grams = 25 / 24 = 1.04 moles

similarly, Molar mass of HCl = 36.5 g / mol

So mass of one mole of HCl = 36.5 g

The moles of  HCl in 20g = 20 / 36.5  = 0.548 moles

Here for one mole of Mg we need two moles of HCl so for 1.04 moles of Mg we will need 2.08 moles of HCl

However we are provided with just 0.548 moles of HCl, hence HCl is the limiting reagent.

0.548 moles of HCl will react with 0.548 /2 moles of Mg

the moles of Mg reacted = 0.274 moles

Moles of excess Mg left = 1.04 - 0.274 = 0.766 moles

Mass of Mg left = moles left X molar mass = 0.766 X 24 = 18.38 g

Answer 2

1) Answer is: limiting reagent is hydrochloric acid.

Balanced chemical reaction: Mg + 2HCl → MgCl₂ + H₂.

m(Mg) = 25 g; mass of magnesium.

n(Mg) = m(Mg) ÷ M(Mg).

n(Mg) = 25 g ÷ 24.3 g/mol.

n(Mg) = 1.03 mol; amount of magnesium.

m(HCl) = 20 g; mass of hydrochloric acid.

M(HCl) = 36.46 g/mol; molar mass of hydrochloric acid.

n(HCl) = m(HCl) ÷ M(HCl).

n(HCl) = 20 g ÷ 36.46 g/mol.

n(HCl) = 0.55 mol; amount of hydrochloric acid.

2) From balanced chemical reaction: n(Mg) : n(HCl) = 1 : 2.

For 1.03 moles of magnesium 2.06 moles of hydrochloric acid is needed.

n(Mg) = n(HCl) ÷ 2.

n(Mg) = 0.55 mol ÷2.

n(Mg) = 0.225 mol.

m(Mg) = 0.225 mol · 24.3 g/mol.

m(Mg) = 6.66 g; mass of magnesium that reacts.

Δm(Mg) = 25 g - 6.66 g.

Δm(Mg) = 18.33 g; the excess reagent remains.