Mole Mass Worksheet Solving Chemical Equations | PDF - Free Printable

Let’s solve this step by step.

First, we need to balance the chemical equation:

Unbalanced:
__ Al + __ Cl₂ → __ AlCl₃

Aluminum (Al) is a single atom on the left, and in AlCl₃ it’s also one Al per formula unit — so that part is okay for now.

Chlorine: On the right, each AlCl₃ has 3 Cl atoms. On the left, Cl₂ has 2 Cl atoms per molecule.

We need to make the number of Cl atoms equal on both sides.

Find least common multiple of 2 and 3 → that’s 6.

So, let’s put 2 in front of AlCl₃ → gives us 2 × 3 = 6 Cl atoms on the right.

To get 6 Cl atoms on the left, we need 3 Cl₂ molecules (because 3 × 2 = 6).

Now check aluminum: Right side has 2 Al (from 2 AlCl₃), so left side needs 2 Al.

Balanced equation:
2 Al + 3 Cl₂ → 2 AlCl₃

✔ Now we can use this balanced equation for all parts.

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Part a:
“2 moles of Al will react with ______ mole(s) of Cl₂ to produce ______ mole(s) of AlCl₃.”

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

So if we have exactly 2 moles of Al, then:

→ It reacts with 3 moles of Cl₂
→ Produces 2 moles of AlCl₃

Answer for a:
3, 2

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Part b:
“How many grams of AlCl₃ will be produced if 2.50 moles of Al react?”

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

Now find molar mass of AlCl₃:

Al = 27.0 g/mol
Cl = 35.5 g/mol → Cl₃ = 3 × 35.5 = 106.5 g/mol
Total = 27.0 + 106.5 = 133.5 g/mol

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

Round appropriately? The given value (2.50) has 3 significant figures → answer should have 3 sig figs.

333.75 → 334 g

Wait — let me double-check multiplication:

2.50 × 133.5

Break it down:
2 × 133.5 = 267
0.5 × 133.5 = 66.75
Total = 267 + 66.75 = 333.75 → yes.

Rounded to 3 sig figs: 334 g

But wait — is 133.5 exact? Actually, atomic masses are often taken as:

Al = 26.98 ≈ 27.0
Cl = 35.45 ≈ 35.5 → so 3×35.45 = 106.35; total = 26.98 + 106.35 = 133.33 g/mol

Hmm — but in most high school problems, they use:

Al = 27.0, Cl = 35.5 → AlCl₃ = 27.0 + 106.5 = 133.5 g/mol

And since 2.50 has 3 sig figs, and 133.5 has 4, limiting is 3 → so 334 g is fine.

Actually, let’s recalculate using more precise values just to be safe:

Standard atomic weights:

Al = 26.9815
Cl = 35.453
AlCl₃ = 26.9815 + 3×35.453 = 26.9815 + 106.359 = 133.3405 g/mol

Then 2.50 mol × 133.3405 g/mol = 333.35125 g → still rounds to 333 g if we consider 2.50 has 3 sig figs? Wait no — 333.35125 rounded to 3 sig figs is 333 g? But 333 has 3 sig figs, and 333.35125 is closer to 333 than 334? No — rounding rules: look at fourth digit.

333.35125 → first three digits: 333, next digit is 3 → less than 5 → round down → 333 g?

Wait — I think I made a mistake earlier.

If we use 133.5 g/mol (as commonly taught):

2.50 × 133.5 = ?

Let me compute again:

133.5 × 2 = 267
133.5 × 0.5 = 66.75
Sum: 267 + 66.75 = 333.75 → which is 333.75

Now, 333.75 with 3 significant figures: look at the third digit is 3 (hundreds place? No — 333.75 — the first digit is 3 (hundreds), second is 3 (tens), third is 3 (units). So 333.75 — to 3 sig figs: look at the fourth digit, which is 7 (tenths place? Wait no — let's write it properly.

Number: 333.75

Significant figures: all non-zero digits are significant. Zeros between or after decimal may be.

Here, 333.75 has 5 sig figs.

We want to round to 3 sig figs because 2.50 has 3.

So, 333.75 → first three digits: 333, then the next digit is 7 (which is >5), so we round up the last 3 to 4 → becomes 334.

Yes! Because 333.75 rounded to 3 significant figures:

Think of it as 3.3375 × 10² → round to 3 sig figs → 3.34 × 10² = 334

So 334 g is correct.

Final answer for b: 334 g

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Part 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 (using standard classroom values)

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

Calculate: 12.3 ÷ 133.5

≈ 0.092134... mol

Keep more digits for intermediate: 12.3 / 133.5 = 123/1335 = simplify? Or just calculate:

133.5 × 0.092 = 133.5 × 0.09 = 12.015, 133.5 × 0.002 = 0.267 → total 12.282 → close to 12.3

Difference: 12.3 - 12.282 = 0.018 → so add 0.018 / 133.5 ≈ 0.000135 → total ≈ 0.092135 mol

So moles AlCl₃ = 12.3 / 133.5 = 0.0921348... ≈ 0.0921 mol (keep 3 sig figs? 12.3 has 3, 133.5 has 4 → so 3 sig figs → 0.0921 mol)

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

So mole ratio: Cl₂ : AlCl₃ = 3 : 2

Therefore, moles Cl₂ needed = (3/2) × moles AlCl₃

= (3/2) × 0.0921348 ≈ 1.5 × 0.0921348 = 0.1382022 mol

With 3 sig figs: 0.138 mol

Check calculation:

12.3 / 133.5 = ?

Do division: 12.3 ÷ 133.5 = 1230 ÷ 13350 = better: 12.3 / 133.5 = 123/1335

Divide numerator and denominator by 3: 41 / 445

41 ÷ 445 ≈ 0.0921348

Then × 1.5 = 0.1382022 → yes.

So 0.138 moles of Cl₂

Answer for c: 0.138 mol

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Part d:
“How many grams of aluminum will react with 3.4 moles of chlorine?”

Note: “chlorine” here means Cl₂, since it’s diatomic.

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

So moles of Al needed = (2/3) × moles of Cl₂

= (2/3) × 3.4 = ?

3.4 × 2 = 6.8
6.8 ÷ 3 = 2.2666... mol Al

Now convert to grams.

Molar mass Al = 27.0 g/mol

Mass = moles × molar mass = 2.2666... × 27.0

First, 2.2666 × 27

Or: (2/3)*3.4 = 6.8/3 = 34/15 mol Al (since 3.4 = 34/10 = 17/5, so (2/3)*(17/5) = 34/15)

34/15 = 2.2666... mol

Now 34/15 × 27.0 = (34 × 27.0) / 15

27.0 / 15 = 1.8

So 34 × 1.8 = ?

30 × 1.8 = 54
4 × 1.8 = 7.2
Total = 61.2 g

Since 3.4 has 2 sig figs, answer should have 2 sig figs.

61.2 → rounded to 2 sig figs is 61 g

Is that right? 3.4 has two sig figs, so yes.

But let me confirm:

(2/3) × 3.4 = 2.2666... mol Al

2.2666 × 27.0 = let's compute:

2 × 27 = 54
0.2666 × 27 ≈ 0.2666 × 20 = 5.332, 0.2666 × 7 = 1.8662 → total ≈ 7.1982
So 54 + 7.1982 = 61.1982 ≈ 61.2 g → with 2 sig figs → 61 g

Yes.

Answer for d: 61 g

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Part 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 = 17 ÷ 27.0 ≈ 0.62963 mol

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

Thus, moles AlCl₃ produced = same as moles Al reacted = 0.62963 mol

Now, 17 g has 2 sig figs, 27.0 has 3 → so limit to 2 sig figs.

0.62963 → to 2 sig figs: 0.63 mol

Because 0.62963 — first two non-zero digits are 6 and 2, but since it's less than 1, leading zero doesn't count, so 6 and 3? Wait:

0.62963 — significant figures: the '6' is first sig fig, '2' is second, '9' is third. To round to 2 sig figs, look at third digit: 9 ≥ 5, so round up the 2 to 3 → 0.63

Yes.

So 0.63 moles

Answer for e: 0.63 mol

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Now, compiling all answers:

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

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