Worksheet on basic stoichiometry with problems involving molar mass and chemical reactions.

Let’s solve each problem step by step. We’ll go one at a time, carefully checking our math.

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Problem 1: Convert grams to moles (and vice versa)

We’re given masses and need to find moles — or sometimes the reverse. To do this, we use molar mass from the periodic table.

Molar mass = sum of atomic masses in the compound (in g/mol)

Formula:
> Moles = Mass (g) ÷ Molar Mass (g/mol)
> Mass (g) = Moles × Molar Mass (g/mol)

Let’s calculate each:

(a) 23.5 g sodium chloride → ? moles

NaCl: Na = 23.0, Cl = 35.5 → Molar mass = 58.5 g/mol
Moles = 23.5 ÷ 58.5 ≈ 0.402 mol

(b) 0.778 g calcium nitrate → ? moles

Ca(NO₃)₂: Ca=40, N=14×2=28, O=16×6=96 → Total = 40+28+96 = 164 g/mol
Moles = 0.778 ÷ 164 ≈ 0.00474 mol

(c) 0.995 g potassium cyanide → ? moles

KCN: K=39, C=12, N=14 → Total = 65 g/mol
Moles = 0.995 ÷ 65 ≈ 0.0153 mol

(d) 15.212 g glucose (C₆H₁₂O₆) → ? moles

C₆H₁₂O₆: C=12×6=72, H=1×12=12, O=16×6=96 → Total = 180 g/mol
Moles = 15.212 ÷ 180 ≈ 0.0845 mol

(e) 100.791 g copper(II) permanganate → ? moles

Cu(MnO₄)₂: Cu=63.5, Mn=55×2=110, O=16×8=128 → Total = 63.5+110+128 = 301.5 g/mol
Moles = 100.791 ÷ 301.5 ≈ 0.334 mol

Now convert moles to grams for parts f–j:

(f) 0.25 g silver sulfate → ? moles? Wait — no! The question says “convert into corresponding number of moles” for all. But part f says “0.25 g silver sulfate” — so same as above.

Wait — looking back: Part 1 says “Convert each of the following units into its corresponding number in grams.” That must be a typo — because then it lists masses and asks for moles? Actually, re-reading:

Actually, the worksheet says:
“Part 1: Convert each of the following units into its corresponding number in grams.”
But then lists things like “23.5 g sodium chloride” — that doesn’t make sense. Probably meant:
→ For (a)-(e): convert grams to moles
→ For (f)-(j): convert moles to grams

Looking at (f): “0.25 g silver sulfate” — again, if it’s already in grams, why convert? Hmm.

Wait — let me check original text again:

It says:
“Part 1: Convert each of the following units into its corresponding number in grams.”

Then lists:
(a) 23.5 g sodium chloride → ??? This is confusing.

Actually, I think there’s a misprint. Looking at standard worksheets, usually:

Part 1 has two sections:
- First half: given mass → find moles
- Second half: given moles → find mass

And indeed, (f) says “0.25 g silver sulfate” — but that can’t be right if we’re converting TO grams.

Wait — perhaps the instruction is wrong. Let me look at what makes sense.

Actually, looking at (f): “0.25 g silver sulfate” — maybe it’s supposed to be “0.25 moles”? Because otherwise it’s redundant.

Similarly, (g) “0.0045 moles barium chloride” — that makes sense to convert to grams.

I think the instruction “into its corresponding number in grams” applies only to (f)-(j), and (a)-(e) are meant to be converted to moles.

That matches typical worksheet structure.

So let’s assume:

For (a)-(e): convert given mass → moles
For (f)-(j): convert given moles → grams

Yes, that fits.

So continuing:

(f) 0.25 moles silver sulfate → ? grams

Ag₂SO₄: Ag=108×2=216, S=32, O=64 → Total = 312 g/mol
Mass = 0.25 × 312 = 78.0 g

(g) 0.0045 moles barium chloride → ? grams

BaCl₂: Ba=137, Cl=35.5×2=71 → Total = 208 g/mol
Mass = 0.0045 × 208 = 0.936 g

(h) 1.8 × 10⁻² moles hydrochloric acid → ? grams

HCl: H=1, Cl=35.5 → 36.5 g/mol
Mass = 0.018 × 36.5 = 0.657 g

(i) 0.015 moles hydrogen sulfide → ? grams

H₂S: H=2, S=32 → 34 g/mol
Mass = 0.015 × 34 = 0.51 g

(j) 0.21 moles sodium bromate → ? grams

NaBrO₃: Na=23, Br=80, O=48 → Total = 151 g/mol
Mass = 0.21 × 151 = 31.71 g

Okay, Part 1 done.

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Part 2: Moles ←→ Number of Particles Conversions

Use Avogadro’s number: 6.022 × 10²³ particles per mole

Formulas:
> Particles = Moles × 6.022e23
> Moles = Particles ÷ 6.022e23

(a) 0.0065 moles phosphorus trichloride → ? molecules

Particles = 0.0065 × 6.022e23 = 3.9143 × 10²¹ molecules

(b) 1.2 × 10²³ molecules dinitrogen pentoxide → ? moles

Moles = 1.2e23 ÷ 6.022e23 ≈ 0.1993 mol

(c) 1.8 × 10⁻²¹ atoms helium → ? moles

Moles = 1.8e-21 ÷ 6.022e23 = 2.989 × 10⁻⁵ mol ← This seems extremely small — double-check exponent.

Wait: 1.8 × 10⁻²¹ divided by 6.022 × 10²³ = (1.8 / 6.022) × 10^(-21-23) = 0.2989 × 10⁻⁴⁴ = 2.989 × 10⁻⁴⁵ mol

Yes, correct — though very tiny.

(d) 4.5 × 10²³ molecules glucose → ? moles

Moles = 4.5e23 ÷ 6.022e23 ≈ 0.747 mol

(e) 8.11 × 10²² atoms krypton → ? moles

Moles = 8.11e22 ÷ 6.022e23 ≈ 0.1347 mol

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Part 3: Grams ↔ Grams Stoichiometry Problems

We use balanced equations and molar ratios.

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Problem 1:

Equation:
2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

Given: 144 g water produced → find various things.

First, molar masses:
- H₂O = 18 g/mol
- C₈H₁₈ = 114 g/mol (C=12×8=96, H=1×18=18 → 114)
- O₂ = 32 g/mol
- CO₂ = 44 g/mol

From equation: 18 moles H₂O come from 2 moles C₈H₁₈ and 25 moles O₂, produce 16 moles CO₂

Step 1: Find moles of H₂O produced

Moles H₂O = 144 g ÷ 18 g/mol = 8.0 moles

Now use ratios:

(a) How many moles of water formed? → Already found: 8.0 moles

(b) How many moles of octane burned?

From equation: 18 mol H₂O ← 2 mol C₈H₁₈
So, 8.0 mol H₂O × (2 mol C₈H₁₈ / 18 mol H₂O) = 16/18 = 0.888... mol ≈ 0.889 mol

(c) How many moles of oxygen used up?

18 mol H₂O ← 25 mol O₂
So, 8.0 × (25/18) = 200/18 ≈ 11.111 mol

(d) How many grams of octane was put in?

Moles octane = 0.8889 mol
Mass = 0.8889 × 114 ≈ 101.33 g

(e) How much carbon dioxide was made?

18 mol H₂O → 16 mol CO₂
So, 8.0 mol H₂O × (16/18) = 128/18 ≈ 7.111 mol CO₂
Mass = 7.111 × 44 ≈ 312.89 g

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Problem 2:

Equation:
2 NaOH + H₂SO₄ → 2 H₂O + Na₂SO₄

Given: 500 g NaOH → how many grams Na₂SO₄?

Molar masses:
- NaOH = 40 g/mol (Na=23, O=16, H=1)
- Na₂SO₄ = 142 g/mol (Na=46, S=32, O=64)

From equation: 2 mol NaOH → 1 mol Na₂SO₄

Moles NaOH = 500 ÷ 40 = 12.5 mol

Moles Na₂SO₄ = 12.5 ÷ 2 = 6.25 mol

Mass Na₂SO₄ = 6.25 × 142 = 887.5 g

Answer given in worksheet: 887.5g Na₂SO₄ — matches.

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Problem 3:

Equation:
Pb(SO₄)₂ + 4 LiNO₃ → Pb(NO₃)₄ + 2 Li₂SO₄

Given: want 250 g lithium sulfate (Li₂SO₄), with excess lead(IV) sulfate → how many grams LiNO₃ needed?

Molar masses:
- Li₂SO₄: Li=7×2=14, S=32, O=64 → 110 g/mol
- LiNO₃: Li=7, N=14, O=48 → 69 g/mol

From equation: 2 mol Li₂SO₄ ← 4 mol LiNO₃

So ratio: 4 mol LiNO₃ : 2 mol Li₂SO₄ → simplify to 2:1

Moles Li₂SO₄ desired = 250 ÷ 110 ≈ 2.2727 mol

Moles LiNO₃ needed = 2 × 2.2727 = 4.5454 mol

Mass LiNO₃ = 4.5454 × 69 ≈ 313.63 g

Answer given: 313.6 g LiNO₃ — matches.

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Problem 4:

Equation:
Fe₂O₃ + 3 H₂ → 2 Fe + 3 H₂O

Given: 14.1 grams Fe₂O₃ → how many grams Fe can be made?

Molar masses:
- Fe₂O₃: Fe=56×2=112, O=48 → 160 g/mol
- Fe: 56 g/mol

From equation: 1 mol Fe₂O₃ → 2 mol Fe

Moles Fe₂O₃ = 14.1 ÷ 160 = 0.088125 mol

Moles Fe = 2 × 0.088125 = 0.17625 mol

Mass Fe = 0.17625 × 56 = 9.87 g

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Now, compiling final answers clearly.

Final Answer:

Part 1:
(a) 0.402 mol
(b) 0.00474 mol
(c) 0.0153 mol
(d) 0.0845 mol
(e) 0.334 mol
(f) 78.0 g
(g) 0.936 g
(h) 0.657 g
(i) 0.51 g
(j) 31.71 g

Part 2:
(a) 3.91 × 10²¹ molecules
(b) 0.199 mol
(c) 2.99 × 10⁻⁴⁵ mol
(d) 0.747 mol
(e) 0.135 mol

Part 3:
1) (a) 8.0 mol H₂O
(b) 0.889 mol C₈H₁₈
(c) 11.1 mol O₂
(d) 101 g C₈H₁₈
(e) 313 g CO₂
2) 887.5 g Na₂SO₄
3) 313.6 g LiNO₃
4) 9.87 g Fe