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25 grams of Mg react with 20 grams of HCl.
1) Identify the limiting reagent.
2) How much of the excess reagent remains?

Please try an explain the best you can, not just the answer as I'm a bit confused

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znk

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 Hâ‚‚:1 mol Mg

Moles of H₂ = 1.03 mol Mg × 1mol H₂/1 mol Mg

Moles of Hâ‚‚ = 1.03 mol Hâ‚‚

From HCl:

The molar ratio is 1 mol Hâ‚‚:2 mol HCl

Moles of H₂ = 0.549 mol HCl × 1 mol H₂/2 mol HCl

Moles of Hâ‚‚ = 0.274 mol Hâ‚‚

The limiting reactant is HCl because it gives the smaller amount of Hâ‚‚.

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

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