Answer:
1) Mg; 2)18 g
Step-by-step explanation:
1) Identify the limiting reactant
We have the masses of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
Step 1. Gather all the information in one place with molar masses above the formulas and masses below them.
M_r: 24.30 36.46
Mg + 2HCl ⟶ MgCl₂ + H₂
Mass/g: 25 20
Step 2. Calculate the moles of each reactant
Moles of Mg = 25 g × 1 mol/24.30 g
Moles of Mg = 1.03 mol Mg
Moles of HCl = 20 g× 1mol/36.46 g
Moles of HCl = 0.549 mol HCl
Step 3. Identify the limiting reactant
Calculate the moles of H₂ we can obtain from each reactant.
From Mg:
The molar ratio is 1 mol H₂:1 mol Mg
Moles of H₂ = 1.03 mol Mg × 1mol H₂/1 mol Mg
Moles of H₂ = 1.03 mol H₂
From HCl:
The molar ratio is 1 mol H₂:2 mol HCl
Moles of H₂ = 0.549 mol HCl × 1 mol H₂/2 mol HCl
Moles of H₂ = 0.274 mol H₂
The limiting reactant is HCl because it gives the smaller amount of H₂.
The excess reactant is Mg.
2) Calculate the mass of Mg remaining
Step 1. Calculate the moles of Mg reacted
The molar ratio is 1 mol Mg: 2 mol HCl
Moles of Mg reacted = 0.549 mol HCl × 1 mol Mg/2 mol HCl
Moles of Mg reacted = 0.274 mol Mg
Step 2. Calculate the mass of Mg reacted
Mass of Mg reacted = 0.274 mol Mg × 24.30 g Mg/1 mol Mg
Mass of Mg reacted = 6.66 g Mg
Step 3. Calculate the mass of Mg remaining
Mass remaining = original mass – mass reacted
Mass remaining = (25 – 6.66) g Mg
Mass remaining = 18 g Mg