Step-by-step solutions to stoichiometry problems involving mass, moles, and molecules in chemical reactions.

It looks like you've provided a list of five stoichiometry problems, each involving chemical reactions and calculations based on mole ratios, molar masses, and Avogadro's number. Let's go through each problem step by step, verify the calculations, and explain the reasoning.

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Problem 1:

Reaction:
$$ 2\text{HI} + \text{Mg} \rightarrow \text{MgI}_2 + \text{H}_2 $$

Given: 15.0 g HI
Find: Number of molecules of H₂ produced

Step-by-step solution:

1. Molar mass of HI = 1.0 (H) + 126.9 (I) ≈ 128 g/mol
(Note: Actually, Iodine is ~126.9, so 1 + 126.9 = 127.9 g/mol — but here it’s approximated as 128 g/mol.)

2. Moles of HI = $ \frac{15.0\ \text{g}}{128\ \text{g/mol}} = 0.1172\ \text{mol} $

3. From balanced equation:
2 mol HI → 1 mol H₂
So moles of H₂ = $ \frac{1}{2} \times 0.1172 = 0.0586\ \text{mol} $

4. Number of molecules of H₂ =
$ 0.0586\ \text{mol} \times 6.02 \times 10^{23}\ \text{molecules/mol} $
= $ 3.53 \times 10^{22} $ molecules

✔ Correct!
Answer: $ 3.53 \times 10^{22} $ molecules of H₂

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Problem 2:

Reaction:
$$ 2\text{Ca} + \text{O}_2 \rightarrow 2\text{CaO} $$

Given: 12.0 g O₂
Find: Mass of Ca required

Solution:

1. Molar mass O₂ = 32.0 g/mol
Moles of O₂ = $ \frac{12.0}{32.0} = 0.375\ \text{mol} $

2. From reaction:
1 mol O₂ reacts with 2 mol Ca
So moles of Ca = $ 2 \times 0.375 = 0.750\ \text{mol} $

3. Molar mass Ca = 40.1 g/mol
Mass of Ca = $ 0.750 \times 40.1 = 30.1\ \text{g} $

✔ Correct!
Answer: 30.1 g Ca

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Problem 3:

Reaction:
$$ \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3 $$

Given: 8.0 g AgCl precipitated
Find: Mass of NaCl solution needed (assuming 100% yield and 9.0 g NaCl per 100 g solution)

Step-by-step:

1. Molar mass AgCl = 107.9 (Ag) + 35.5 (Cl) = 143.4 g/mol
(Here, they used 143 g/mol — close enough.)

2. Moles of AgCl = $ \frac{8.0}{143} = 0.0560\ \text{mol} $

3. From reaction: 1 mol AgCl ←→ 1 mol NaCl
So moles of NaCl needed = 0.0560 mol

4. Molar mass NaCl = 58.5 g/mol
Mass of NaCl = $ 0.0560 \times 58.5 = 3.27\ \text{g} $

5. But the solution contains 9.0 g NaCl per 100 g solution
So mass of solution = $ \frac{100\ \text{g soln}}{9.0\ \text{g NaCl}} \times 3.27\ \text{g NaCl} = 36.3\ \text{g soln} $

Wait — but in your calculation, you wrote:

> $ \frac{58.5\ \text{g NaCl}}{1\ \text{mol NaCl}} \times \frac{100\ \text{g NaCl soln}}{9.0\ \text{g NaCl}} = 81.8\ \text{g NaCl soln} $

This seems to be incorrect because:

- You’re multiplying 58.5 × (100/9.0) = 649.4 → then multiplying that by 0.056? No — you skipped steps.

Let’s re-express correctly:

We have:
- Moles of AgCl = 8.0 / 143 = 0.0560 mol
- So need 0.0560 mol NaCl
- Mass of NaCl = 0.0560 × 58.5 = 3.27 g
- This NaCl comes from a solution that is 9.0 g NaCl per 100 g solution

So:
$$
\text{Mass of solution} = \frac{100\ \text{g soln}}{9.0\ \text{g NaCl}} \times 3.27\ \text{g NaCl} = 36.3\ \text{g soln}
$$

But you wrote 81.8 g — which suggests an error.

Wait — let's check what you did:

You wrote:
> $ 8.0\ \text{g AgCl} \times \frac{1\ \text{mol AgCl}}{143\ \text{g}} \times \frac{1\ \text{mol NaCl}}{1\ \text{mol AgCl}} \times \frac{58.5\ \text{g NaCl}}{1\ \text{mol NaCl}} \times \frac{100\ \text{g NaCl soln}}{9.0\ \text{g NaCl}} $

That’s:
$$
8.0 \times \frac{1}{143} \times 1 \times 58.5 \times \frac{100}{9.0}
= 8.0 \times \frac{58.5 \times 100}{143 \times 9.0}
$$

Calculate:
- Numerator: $ 8.0 \times 58.5 \times 100 = 46,800 $
- Denominator: $ 143 \times 9.0 = 1,287 $
- Result: $ \frac{46,800}{1,287} \approx 36.36\ \text{g} $

So you should get ~36.4 g, not 81.8 g.

But you wrote 81.8 g NaCl soln — this is twice as much.

Possibility: Did you use 16.0 g AgCl instead of 8.0?

Check:
If it were 16.0 g AgCl:
- $ 16.0 \times \frac{58.5 \times 100}{143 \times 9.0} = 72.7 $ — still not 81.8.

Wait — maybe you meant something else?

Alternatively, perhaps the solution concentration is misread.

Wait — look at your expression:
> $ \frac{100\ \text{g NaCl soln}}{9.0\ \text{g NaCl}} $

That means 9.0 g NaCl per 100 g solution, so 9% w/w.

So for 3.27 g NaCl, solution mass = $ \frac{100}{9.0} \times 3.27 = 36.3\ \text{g} $

But you got 81.8 g — that’s double.

Wait — could it be that you used 1 mol AgCl → 2 mol NaCl? No — the reaction is 1:1.

Wait — let’s recalculate your final expression:

You wrote:
> $ 8.0 \times \frac{1}{143} \times \frac{1}{1} \times 58.5 \times \frac{100}{9.0} $

= $ 8.0 \times \frac{58.5 \times 100}{143 \times 9.0} $

= $ 8.0 \times \frac{5850}{1287} $

= $ 8.0 \times 4.545 = 36.36\ \text{g} $

So your final answer of 81.8 g is incorrect.

But you wrote 81.8 g NaCl soln — likely a mistake.

✔ Correct Answer: 36.4 g NaCl solution

✘ Your answer (81.8 g) is wrong — probably due to a calculation error or typo.

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Problem 4:

Reaction:
$$ 2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2 $$

Given: 0.320 mol O₂ produced
Find: Mass of H₂O₂ decomposed

Solution:

From reaction:
1 mol O₂ ←→ 2 mol H₂O₂

So moles of H₂O₂ = $ 2 \times 0.320 = 0.640\ \text{mol} $

Molar mass H₂O₂ = 2(1.0) + 2(16.0) = 34.0 g/mol

Mass of H₂O₂ = $ 0.640 \times 34.0 = 21.8\ \text{g} $

✔ Correct!
Answer: 21.8 g H₂O₂

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Problem 5:

Reaction:
$$ 3\text{Cl}_2 + 3\text{O}_2 \rightarrow 2\text{Cl}_2\text{O}_3 $$
Wait — let’s balance it properly.

Actually, the correct formation of Cl₂O₃ is:

$$ 2\text{Cl}_2 + 3\text{O}_2 \rightarrow 2\text{Cl}_2\text{O}_3 $$

Yes — this matches the ratio in your calculation.

So:
3 mol O₂ → 2 mol Cl₂O₃
(From your work: $ \frac{2\ \text{mol Cl}_2\text{O}_3}{3\ \text{mol O}_2} $)

Given: 16.0 g O₂
Find: Mass of Cl₂O₃ formed

Steps:

1. Moles of O₂ = $ \frac{16.0}{32.0} = 0.500\ \text{mol} $

2. From ratio:
$ \frac{2\ \text{mol Cl}_2\text{O}_3}{3\ \text{mol O}_2} $
So moles of Cl₂O₃ = $ \frac{2}{3} \times 0.500 = 0.333\ \text{mol} $

3. Molar mass Cl₂O₃ = 2(35.5) + 3(16.0) = 71.0 + 48.0 = 119.0 g/mol

4. Mass of Cl₂O₃ = $ 0.333 \times 119.0 = 39.7\ \text{g} $

✔ Correct!
Answer: 39.7 g Cl₂O₃

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✔ Summary of Corrections:

| Problem | Your Answer | Correct? | Notes |
|--------|-------------|---------|-------|
| 1 | $ 3.53 \times 10^{22} $ | ✔ | Correct |
| 2 | 30.1 g Ca | ✔ | Correct |
| 3 | 81.8 g NaCl soln | ✘ | Should be 36.4 g — you likely made a calculation error |
| 4 | 21.8 g H₂O₂ | ✔ | Correct |
| 5 | 39.7 g Cl₂O₃ | ✔ | Correct |

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🔍 Fix for Problem 3:

Corrected Calculation:

$$
8.0\ \text{g AgCl} \times \frac{1\ \text{mol AgCl}}{143\ \text{g}} \times \frac{1\ \text{mol NaCl}}{1\ \text{mol AgCl}} \times \frac{58.5\ \text{g NaCl}}{1\ \text{mol NaCl}} \times \frac{100\ \text{g soln}}{9.0\ \text{g NaCl}} = \boxed{36.4\ \text{g NaCl solution}}
$$

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Final Answers:

1. $ 3.53 \times 10^{22} $ molecules H₂
2. 30.1 g Ca
3. 36.4 g NaCl solution (not 81.8 g)
4. 21.8 g H₂O₂
5. 39.7 g Cl₂O₃

Let me know if you want the corrected version formatted neatly!