Mole/Mass Problems worksheet with solved chemistry equations and calculations.

Let’s solve this step by step.

We are told that aluminum (Al) reacts with chlorine (Cl₂) to produce aluminum chloride (AlCl₃). First, we need to write and balance the chemical equation.

Unbalanced:
Al + Cl₂ → AlCl₃

Balance it:

- Aluminum: 1 on left, 1 on right → OK for now.
- Chlorine: 2 on left, 3 on right → not balanced.

To balance chlorine, find least common multiple of 2 and 3 → which is 6.

So, put 3 in front of Cl₂ (gives 6 Cl atoms), and 2 in front of AlCl₃ (also gives 6 Cl atoms).

Now:
Al + 3Cl₂ → 2AlCl₃

But now aluminum is unbalanced — 2 Al on right, only 1 on left. So put 2 in front of Al.

Balanced equation:
2Al + 3Cl₂ → 2AlCl₃

This means:
- 2 moles of Al react with 3 moles of Cl₂ to make 2 moles of AlCl₃.

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Now let’s answer each part.

a. 2 moles of Al will react with ___ mole(s) of Cl₂ to produce ___ mole(s) of AlCl₃.

From the balanced equation:
2 mol Al : 3 mol Cl₂ : 2 mol AlCl₃

So:
→ 2 moles Al react with 3 moles Cl₂ → produces 2 moles AlCl₃.

✔ Answer: 3, 2

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b. How many grams of AlCl₃ will be produced if 2.50 moles of Al react?

From the equation: 2 mol Al → 2 mol AlCl₃
So ratio is 1:1 → 2.50 mol Al → 2.50 mol AlCl₃

Now convert moles of AlCl₃ to grams.

Molar mass of AlCl₃ = Al + 3×Cl = 27.0 g/mol + 3×35.5 g/mol = 27.0 + 106.5 = 133.5 g/mol

Mass = moles × molar mass = 2.50 mol × 133.5 g/mol = 333.75 g

Round to 3 significant figures (since 2.50 has 3 sig figs):
→ 334 g

✔ Answer: 334 g

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c. How many moles of Cl₂ must react to produce 12.3 g of AlCl₃?

First, convert grams of AlCl₃ to moles.

Molar mass AlCl₃ = 133.5 g/mol (as above)

Moles of AlCl₃ = mass / molar mass = 12.3 g / 133.5 g/mol ≈ 0.09213 mol

From balanced equation:
2 mol AlCl₃ ← requires 3 mol Cl₂

So set up proportion:

If 2 mol AlCl₃ come from 3 mol Cl₂,
then 0.09213 mol AlCl₃ comes from ? mol Cl₂

Use ratio:
mol Cl₂ = (3/2) × mol AlCl₃ = 1.5 × 0.09213 ≈ 0.1382 mol

Round to 3 sig figs (12.3 has 3 sig figs):
→ 0.138 mol

✔ Answer: 0.138 mol

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d. How many grams of aluminum will react with 3.4 moles of chlorine?

From balanced equation:
3 mol Cl₂ react with 2 mol Al

So, 3.4 mol Cl₂ → how many mol Al?

Set up ratio:
mol Al = (2/3) × mol Cl₂ = (2/3) × 3.4 ≈ 2.2667 mol Al

Now convert to grams.

Molar mass Al = 27.0 g/mol

Mass = 2.2667 mol × 27.0 g/mol ≈ 61.2 g

(3.4 has 2 sig figs? Wait — 3.4 has 2 sig figs, but often in these problems, we treat it as exact or keep intermediate precision. Let’s check: 3.4 could be 2 sig figs, so answer should have 2 sig figs? But 3.4 might be considered as having 2, but in context, maybe 2.2667 × 27.0 = 61.2009 → round to 61 g? But let’s see what makes sense.)

Actually, 3.4 has two significant figures, so answer should have two.

2.2667 mol × 27.0 g/mol = 61.2009 → rounds to 61 g

But wait — 27.0 has 3 sig figs, 3.4 has 2 → limiting is 2 sig figs.

So → 61 g

But let me double-check calculation:

(2/3)*3.4 = 6.8/3 = 2.2666... mol Al

2.2666... × 27.0 = let's compute:

2.2666 × 27 = 2.2666 × 20 + 2.2666 × 7 = 45.332 + 15.8662 = 61.1982 → yes, ~61.2

With 2 sig figs → 61 g

But sometimes teachers accept 61.2 if they consider 3.4 as precise. However, strictly speaking, 3.4 has 2 sig figs → answer should be 61 g.

Looking back at student’s work — they wrote 61.16g → probably kept more digits. But for accuracy, we’ll go with proper sig figs.

Wait — actually, in many school settings, unless specified, they may expect you to use the numbers as given without strict sig fig rounding. Since 3.4 might be intended as 3.40, but it’s written as 3.4.

I think safest is to report as 61 g (2 sig figs).

But let me check part e — they used 17 g (2 sig figs) and got 0.37 mol (2 sig figs). So consistent.

So here: 3.4 mol → 2 sig figs → answer 61 g.

✔ Answer: 61 g

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e. If 17 grams of aluminum react, how many moles of aluminum chloride will be produced?

First, convert grams Al to moles.

Molar mass Al = 27.0 g/mol

Moles Al = 17 g / 27.0 g/mol ≈ 0.6296 mol

From balanced equation:
2 mol Al → 2 mol AlCl₃ → so 1:1 ratio

Therefore, moles AlCl₃ = moles Al = 0.6296 mol

Now, 17 g has 2 sig figs → so answer should have 2 sig figs.

0.6296 → rounds to 0.63 mol

✔ Answer: 0.63 mol

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Final Answers Summary:

a. 3, 2
b. 334 g
c. 0.138 mol
d. 61 g
e. 0.63 mol

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Final Answer:
a. 3, 2
b. 334 g
c. 0.138 mol
d. 61 g
e. 0.63 mol