Enthalpy Stoichiometry Worksheet featuring problems on heat release and absorption in chemical reactions.

Let's solve each problem step by step using enthalpy stoichiometry principles. We'll use molar masses, stoichiometric ratios, and the given ΔH° values to find the heat released or absorbed.

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1. How much heat will be released when 4.72 g of carbon react with excess O₂?

Reaction:
$$
\text{C} + \text{O}_2 \rightarrow \text{CO}_2; \quad \Delta H^\circ = -393.5 \, \text{kJ}
$$

This means 1 mole of C releases 393.5 kJ of heat.

Step 1: Molar mass of carbon (C) = 12.01 g/mol
Step 2: Moles of C = $ \frac{4.72}{12.01} = 0.393 \, \text{mol} $

Step 3: Heat released = $ 0.393 \, \text{mol} \times 393.5 \, \text{kJ/mol} = 154.6 \, \text{kJ} $

✔ Answer: 154.6 kJ of heat is released.

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2. How much heat will be absorbed when 38.2 g of bromine react with excess H₂?

Reaction:
$$
\text{H}_2 + \text{Br}_2 \rightarrow 2\text{HBr}; \quad \Delta H^\circ = +72.80 \, \text{kJ}
$$

Positive ΔH → endothermic → heat absorbed

1 mole of Br₂ absorbs 72.80 kJ.

Step 1: Molar mass of Br₂ = $ 2 \times 79.90 = 159.8 \, \text{g/mol} $

Step 2: Moles of Br₂ = $ \frac{38.2}{159.8} = 0.239 \, \text{mol} $

Step 3: Heat absorbed = $ 0.239 \times 72.80 = 17.37 \, \text{kJ} $

✔ Answer: 17.37 kJ of heat is absorbed.

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3. How much heat will be released when 1.48 g of chlorine react with excess phosphorus?

Reaction:
$$
2\text{P} + 5\text{Cl}_2 \rightarrow 2\text{PCl}_5; \quad \Delta H^\circ = -886 \, \text{kJ}
$$

ΔH = -886 kJ for 5 moles of Cl₂

So, 5 mol Cl₂ → 886 kJ released

Step 1: Molar mass of Cl₂ = $ 2 \times 35.45 = 70.90 \, \text{g/mol} $

Step 2: Moles of Cl₂ = $ \frac{1.48}{70.90} = 0.02087 \, \text{mol} $

Step 3: Use proportion:

$$
\frac{5 \, \text{mol Cl}_2}{886 \, \text{kJ}} = \frac{0.02087 \, \text{mol}}{x}
\Rightarrow x = \frac{0.02087 \times 886}{5} = 3.69 \, \text{kJ}
$$

✔ Answer: 3.69 kJ of heat is released.

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4. How much heat will be absorbed when 13.7 g of nitrogen react with excess O₂?

Reaction:
$$
\text{N}_2 + \text{O}_2 \rightarrow 2\text{NO}; \quad \Delta H^\circ = +180 \, \text{kJ}
$$

+180 kJ per mole of N₂ → endothermic

Step 1: Molar mass of N₂ = 28.02 g/mol

Step 2: Moles of N₂ = $ \frac{13.7}{28.02} = 0.489 \, \text{mol} $

Step 3: Heat absorbed = $ 0.489 \times 180 = 87.96 \, \text{kJ} $

✔ Answer: 87.96 kJ of heat is absorbed.

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5. Find the heat of reaction for the single displacement reaction in which 2 L of chlorine gas at STP react with sodium bromide.

Reaction:
$$
\text{Cl}_2(g) + 2\text{NaBr}(aq) \rightarrow \text{Br}_2(l) + 2\text{NaCl}(aq); \quad \Delta H^\circ = -100.2 \, \text{kJ}
$$

This ΔH° is for 1 mole of Cl₂.

We are given 2 L of Cl₂ at STP.

At STP, 1 mole = 22.4 L

So, moles of Cl₂ = $ \frac{2}{22.4} = 0.0893 \, \text{mol} $

Heat released = $ 0.0893 \times 100.2 = 8.95 \, \text{kJ} $

But the question says: "Find the heat of reaction" — this might mean the total heat change for the given amount.

Since ΔH° = -100.2 kJ per mole of Cl₂, and we have 0.0893 mol:

$$
q = 0.0893 \times (-100.2) = -8.95 \, \text{kJ}
$$

So, 8.95 kJ of heat is released.

✔ Answer: -8.95 kJ (heat released)

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6. A chemist runs an experiment where:

$$
\text{CrO}_3(s) + \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{CrO}_4(aq); \quad \Delta H^\circ = -5.4 \, \text{kJ}
$$

Energy released = 6.18 kJ (experimentally measured)

We need to find the mass of CrO₃ that reacted.

Step 1: The reaction shows that 1 mole of CrO₃ releases 5.4 kJ of energy.

So, moles of CrO₃ that reacted:

$$
\frac{6.18 \, \text{kJ}}{5.4 \, \text{kJ/mol}} = 1.144 \, \text{mol}
$$

Step 2: Molar mass of CrO₃ = Cr (52.00) + 3×O (3×16.00) = 52.00 + 48.00 = 100.00 g/mol

Step 3: Mass = $ 1.144 \times 100.00 = 114.4 \, \text{g} $

✔ Answer: 114.4 g of CrO₃ reacted.

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✔ Final Answers Summary:

1. 154.6 kJ released
2. 17.37 kJ absorbed
3. 3.69 kJ released
4. 87.96 kJ absorbed
5. -8.95 kJ (heat released)
6. 114.4 g of CrO₃ reacted

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