Stoichiometry Using Molarity Practice Problems

Let’s solve each problem one by one, step by step. We’ll use the balanced chemical equations and molarity concepts to find the answers.

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Problem 1:
Chemical equation:
H₂SO₄ + NaOH → Na₂SO₄ + H₂O
But this is NOT balanced! Let’s balance it first.

Balanced equation:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

We’re told:
- Volume of NaOH = 43.2 mL = 0.0432 L
- Molarity of NaOH = 0.236 M
- Volume of H₂SO₄ = 36.7 mL = 0.0367 L
- Find molarity of H₂SO₄

Step 1: Find moles of NaOH used.
Moles = Molarity × Volume (in liters)
→ Moles NaOH = 0.236 mol/L × 0.0432 L = 0.0101952 mol

Step 2: Use mole ratio from balanced equation.
From H₂SO₄ + 2NaOH → ..., 1 mol H₂SO₄ reacts with 2 mol NaOH.
So, moles H₂SO₄ needed = moles NaOH ÷ 2 = 0.0101952 ÷ 2 = 0.0050976 mol

Step 3: Find molarity of H₂SO₄.
Molarity = moles ÷ volume (L)
→ Molarity H₂SO₄ = 0.0050976 mol ÷ 0.0367 L ≈ 0.1389 M

Rounded to 3 significant figures: 0.139 M

✔ Final Answer for Problem 1: 0.139 M

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Problem 2:
Chemical equation:
NaOH + HCl → H₂O + NaCl
Already balanced! 1:1 ratio.

Given:
- Volume HCl = 36.7 mL = 0.0367 L
- Volume NaOH = 43.2 mL = 0.0432 L
- Molarity NaOH = 0.236 M
- Find molarity of HCl

Step 1: Moles of NaOH = 0.236 × 0.0432 = 0.0101952 mol

Step 2: Since 1:1 ratio, moles HCl = same = 0.0101952 mol

Step 3: Molarity HCl = moles ÷ volume (L) = 0.0101952 ÷ 0.0367 ≈ 0.2778 M

Rounded to 3 sig figs: 0.278 M

✔ Final Answer for Problem 2: 0.278 M

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Problem 3:
Chemical equation:
Al(OH)₃ + HCl → AlCl₃ + H₂O
Not balanced. Balance it:

Al(OH)₃ has 3 OH⁻, so needs 3 H⁺ → 3 HCl
Products: AlCl₃ and 3 H₂O

Balanced:
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O

Given:
- Mass Al(OH)₃ = 5.8 g
- Molarity HCl = 1.2 M
- Find volume of HCl in mL

Step 1: Molar mass of Al(OH)₃
Al = 27.0, O = 16.0×3=48.0, H = 1.0×3=3.0 → Total = 27+48+3 = 78.0 g/mol

Step 2: Moles Al(OH)₃ = mass ÷ molar mass = 5.8 g ÷ 78.0 g/mol ≈ 0.074359 mol

Step 3: From equation, 1 mol Al(OH)₃ needs 3 mol HCl
→ Moles HCl needed = 0.074359 × 3 = 0.223077 mol

Step 4: Volume HCl (L) = moles ÷ molarity = 0.223077 ÷ 1.2 ≈ 0.1859 L

Convert to mL: 0.1859 × 1000 = 185.9 mL

Rounded to 2 sig figs (since 5.8 has 2): 190 mL? Wait — let’s check sig figs.

Actually, 5.8 has 2 sig figs, 1.2 has 2 sig figs → answer should have 2 sig figs.

0.223077 ÷ 1.2 = 0.1859 → 0.19 L = 190 mL

But wait — 5.8 / 78.0 = 0.074359… which we can keep as is for calculation, but final answer rounded to 2 sig figs.

Alternatively, maybe 5.8 is considered 2 sig figs, but often in these problems, they expect you to carry through and round at end.

Let me recalculate precisely:

5.8 / 78.0 = 0.07435897...

×3 = 0.2230769...

÷1.2 = 0.185897... L = 185.897... mL → rounds to 186 mL if using 3 sig figs? But inputs are 5.8 (2 sig figs), 1.2 (2 sig figs). So answer should be 190 mL.

Wait — actually, 5.8 could be interpreted as having 2 sig figs, but sometimes trailing zeros after decimal count. Here, no decimal issue. Standard rule: multiplication/division → least number of sig figs.

5.8 → 2 sig figs
1.2 → 2 sig figs
78.0 → 3 sig figs (but it’s exact molar mass? Usually we treat molar masses as exact for sig fig purposes)

In many textbooks, molar masses are treated as exact, so limiting sig figs come from measured values: 5.8 g and 1.2 M → both 2 sig figs.

So answer: 190 mL

But let’s see what the precise value is: 185.9 mL → rounding to 2 sig figs is 190 mL.

However, some might argue that 5.8 has uncertainty in tenths place, so 186 mL is acceptable. But strictly, 2 sig figs → 1.9 × 10² mL = 190 mL

I think for safety, since 5.8 and 1.2 both have 2 sig figs, we go with 190 mL

But let me double-check with another approach.

Perhaps the problem expects 3 sig figs? Looking back, other problems had 3 sig figs. Maybe 5.8 is meant to be 5.80? No, written as 5.8.

Actually, in many classroom settings, they accept 186 mL here. Let me calculate numerically again:

5.8 / 78 = 0.07435897
×3 = 0.2230769
/1.2 = 0.1858974 L = 185.8974 mL → typically rounded to 186 mL

And since 1.2 has two sig figs, but it's 1.2 not 1.20, still 2 sig figs. However, 186 has three digits — but the first two are certain? This is ambiguous.

Looking at similar problems, often they report to 3 sig figs even if input has 2, especially if intermediate steps allow.

To be safe, I'll go with 186 mL, as it's more precise and commonly accepted.

Wait — let's see the next problems; they use 3 sig figs often. Perhaps here too.

I recall that in stoichiometry, if mass is given as 5.8, it might imply ±0.1, so relative error about 1.7%, so answer should reflect that. 186 mL ± 3 mL or so. So 186 is fine.

But let's stick with calculation: 185.9 mL → 186 mL

✔ Final Answer for Problem 3: 186 mL

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Problem 4:
Chemical equation:
H₂SO₄ + Na₂CO₃ → Na₂SO₄ + H₂O + CO₂
Balance it:

Left: H₂, S, O₄ + Na₂, C, O₃ → total O: 7? Better to balance atoms.

Actually:
H₂SO₄ + Na₂CO₃ → Na₂SO₄ + H₂CO₃
But H₂CO₃ decomposes to H₂O + CO₂, so overall:

H₂SO₄ + Na₂CO₃ → Na₂SO₄ + H₂O + CO₂

Check atoms:
Left: H=2, S=1, O=4+3=7, Na=2, C=1
Right: Na=2, S=1, O=4+1+2=7, H=2, C=1 → balanced!

So 1:1 ratio between H₂SO₄ and Na₂CO₃.

Given:
- Volume H₂SO₄ = 40.0 mL = 0.0400 L
- Mass Na₂CO₃ = 0.364 g
- Find molarity of H₂SO₄

Step 1: Molar mass Na₂CO₃
Na=23.0×2=46.0, C=12.0, O=16.0×3=48.0 → Total = 46+12+48 = 106.0 g/mol

Step 2: Moles Na₂CO₃ = 0.364 g ÷ 106.0 g/mol = 0.00343396 mol

Step 3: Since 1:1 ratio, moles H₂SO₄ = same = 0.00343396 mol

Step 4: Molarity H₂SO₄ = moles ÷ volume (L) = 0.00343396 ÷ 0.0400 = 0.085849 M

Rounded to 3 sig figs (0.364 has 3, 40.0 has 3): 0.0858 M

✔ Final Answer for Problem 4: 0.0858 M

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Problem 5:
Chemical equation:
Ca(OH)₂(s) + HCl(aq) → CaCl₂(aq) + H₂O(l)
Not balanced.

Ca(OH)₂ has 2 OH⁻, so needs 2 H⁺ → 2 HCl
Products: CaCl₂ and 2 H₂O

Balanced:
Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

Given:
- Mass Ca(OH)₂ = 5.00 grams
- Molarity HCl = 0.100 M
- Find volume of HCl in liters

Step 1: Molar mass Ca(OH)₂
Ca=40.1, O=16.0×2=32.0, H=1.0×2=2.0 → Total = 40.1+32.0+2.0 = 74.1 g/mol

Step 2: Moles Ca(OH)₂ = 5.00 g ÷ 74.1 g/mol ≈ 0.067476 mol

Step 3: From equation, 1 mol Ca(OH)₂ requires 2 mol HCl
→ Moles HCl needed = 0.067476 × 2 = 0.134952 mol

Step 4: Volume HCl (L) = moles ÷ molarity = 0.134952 ÷ 0.100 = 1.34952 L

Rounded to 3 sig figs (5.00 and 0.100 both have 3): 1.35 L

✔ Final Answer for Problem 5: 1.35 L

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Problem 6:
Same balanced equation:
Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

Given:
- Mass Ca(OH)₂ = 15.0 grams
- Volume HCl = 75.0 mL = 0.0750 L
- Molarity HCl = 0.500 M
- Find grams of CaCl₂ formed

This is a limiting reactant problem! We need to find which reactant limits the product.

Step 1: Find moles of each reactant.

Moles Ca(OH)₂ = mass ÷ molar mass = 15.0 g ÷ 74.1 g/mol ≈ 0.202429 mol

Moles HCl = M × V = 0.500 mol/L × 0.0750 L = 0.0375 mol

Step 2: Use stoichiometry to see how much product each can make.

From equation:
1 mol Ca(OH)₂ produces 1 mol CaCl₂
2 mol HCl produce 1 mol CaCl₂

So, from Ca(OH)₂: max CaCl₂ = 0.202429 mol

From HCl: max CaCl₂ = 0.0375 mol HCl × (1 mol CaCl₂ / 2 mol HCl) = 0.01875 mol

Since HCl produces less CaCl₂, HCl is limiting reactant.

Step 3: Calculate mass of CaCl₂ from limiting reactant.

Molar mass CaCl₂: Ca=40.1, Cl=35.5×2=71.0 → Total = 40.1+71.0 = 111.1 g/mol

Mass CaCl₂ = moles × molar mass = 0.01875 mol × 111.1 g/mol ≈ ?

Calculate: 0.01875 × 111.1 = let's compute:

0.01875 × 100 = 1.875
0.01875 × 11.1 = 0.01875 × 10 = 0.1875, 0.01875 × 1.1 = 0.020625 → total 0.1875 + 0.020625 = 0.208125
Better: 0.01875 × 111.1 = (1875/100000) × 111.1 = but easier:

0.01875 × 111.1 = 0.01875 × (100 + 11.1) = 1.875 + 0.208125 = 2.083125? That can't be right because 0.01875 × 100 = 1.875, yes, but 0.01875 × 11.1:

0.01875 × 10 = 0.1875
0.01875 × 1 = 0.01875
0.01875 × 0.1 = 0.001875
So 0.1875 + 0.01875 = 0.20625 + 0.001875 = 0.208125
Total: 1.875 + 0.208125 = 2.083125 g? But that seems low.

Wait, 0.01875 mol × 111.1 g/mol:

Do 1875 × 1111 / 100000 × 10? Better use calculator in mind.

0.01875 × 111.1 = ?

First, 0.02 × 111.1 = 2.222
But 0.01875 is 0.02 - 0.00125
0.00125 × 111.1 = 0.138875
So 2.222 - 0.138875 = 2.083125 g

Yes, approximately 2.083 g

Now, sig figs: inputs are 15.0 (3 sig figs), 75.0 (3), 0.500 (3) → so answer should have 3 sig figs.

2.083 g → 2.08 g? But 2.083 rounded to 3 sig figs is 2.08 g.

Wait, 2.083 has four digits, but the value is 2.083, so to three sig figs: look at fourth digit is 3<5, so 2.08 g

But let's confirm the multiplication:

0.01875 × 111.1

Compute exactly:
1875/100000 × 1111/10 = (1875 × 1111) / 1,000,000

Easier: 0.01875 × 111.1 = 0.01875 × (111 + 0.1) = 0.01875×111 + 0.01875×0.1

0.01875×100=1.875, 0.01875×11=0.20625, so 1.875+0.20625=2.08125 for 111? No:

0.01875 × 111 = 0.01875 × (100 + 10 + 1) = 1.875 + 0.1875 + 0.01875 = 2.08125

Then 0.01875 × 0.1 = 0.001875

Total: 2.08125 + 0.001875 = 2.083125 g

Yes.

So 2.083125 g → to three significant figures: 2.08 g

But is that correct? 2.08 has three sig figs, yes.

Some might say 2.08 g, but let's see if we should consider the moles.

Moles HCl = 0.500 × 0.0750 = 0.0375 mol (exactly, since 0.500 and 0.0750 both have three sig figs, product has three: 0.0375)

Then moles CaCl₂ = 0.0375 / 2 = 0.01875 mol (still three sig figs? 0.01875 has four, but from division by 2, which is exact, so still three sig figs from 0.0375)

0.0375 has three sig figs, divided by 2 (exact) → 0.01875, which we can consider as 0.0188 mol if rounded to three sig figs? No, 0.0375 / 2 = 0.01875, and since 2 is exact, it keeps the precision of 0.0375, which is three sig figs, so 0.0188 mol? But 0.0375 has three sig figs, meaning ±0.0005, so 0.01875 ±0.00025, so when multiplying by molar mass, we get uncertainty.

Typically, we carry the calculation and round at end.

Molar mass CaCl₂ = 111.1 g/mol (four sig figs, but often taken as exact for this purpose)

So mass = 0.01875 × 111.1 = 2.083125 g

With input data having three sig figs, answer should be 2.08 g

But let's check online or standard practice. In many cases, they would report 2.08 g.

Note: 0.01875 mol × 111.1 g/mol = let's do 1875e-5 * 111.1 = but anyway.

I think 2.08 g is correct.

But wait — 15.0 g Ca(OH)₂ was given, but it's in excess, so doesn't affect the product amount since HCl is limiting.

Yes.

So final answer: 2.08 g

✔ Final Answer for Problem 6: 2.08 g

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Now, compiling all final answers:

Final Answer:
Problem 1: 0.139 M
Problem 2: 0.278 M
Problem 3: 186 mL
Problem 4: 0.0858 M
Problem 5: 1.35 L
Problem 6: 2.08 g