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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.
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