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.
<em>Step 1</em>. 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
<em>Step 2.</em> Calculate the <em>moles of each reactant </em>
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. <em>Identify the limiting reactant </em>
Calculate the moles of H₂ we can obtain from each reactant.
<em>From Mg:
</em>
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
<em>From HCl:
</em>
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 <em>excess reactant is Mg</em>.
2) Calculate the mass of Mg remaining
<em>Step 1</em>. Calculate the <em>moles of Mg reacted</em>
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
<em>Step 2</em>. Calculate the <em>mass of Mg reacted
</em>
Mass of Mg reacted = 0.274 mol Mg × 24.30 g Mg/1 mol Mg
Mass of Mg reacted = 6.66 g Mg
<em>Step 3</em>. Calculate the <em>mass of Mg remaining</em>
Mass remaining = original mass – mass reacted
Mass remaining = (25 – 6.66) g Mg
Mass remaining = 18 g Mg