Question

consider the following decomposition reaction:
2 hbr(g) + 72.4 kj → h₂(g) + br₂(l)
calculate the mass (in grams) of br₂ produced when 5.76×10⁴ kj is consumed in this reaction. report your answer in scientific notation. be sure your answer has the correct number of significant figures.
□ g br₂

Explanation:

Step1: Find moles of energy consumed

From the reaction, 72.4 kJ energy is consumed to produce 1 mole of \(Br_2\) (since the stoichiometric coefficient of \(Br_2\) is 1). Let \(n\) be the moles of \(Br_2\) produced when \(5.76\times10^{4}\) kJ is consumed. Using the ratio: \(\frac{n}{5.76\times10^{4}\text{ kJ}}=\frac{1\text{ mol}}{72.4\text{ kJ}}\), so \(n = \frac{5.76\times10^{4}\text{ kJ}}{72.4\text{ kJ/mol}}\approx795.58\text{ mol}\).

Step2: Calculate molar mass of \(Br_2\)

Molar mass of \(Br\) is 79.904 g/mol, so for \(Br_2\), it is \(2\times79.904 = 159.808\) g/mol.

Step3: Find mass of \(Br_2\)

Mass \(m = n\times\text{molar mass}\). Substituting \(n = 795.58\) mol and molar mass = 159.808 g/mol: \(m=795.58\text{ mol}\times159.808\text{ g/mol}\approx1.271\times10^{5}\text{ g}\). Checking significant figures: 5.76 (3 sig figs), 72.4 (3 sig figs), so the answer should have 3 sig figs. So \(m\approx1.27\times10^{5}\) g (or more accurately, let's recalculate with more precision:

First, \(n=\frac{5.76\times10^{4}}{72.4}=\frac{57600}{72.4}\approx795.5801\) mol.

Then \(m = 795.5801\times159.808\approx795.5801\times160 - 795.5801\times0.192\approx127292.816 - 152.751\approx127140.065\approx1.27\times10^{5}\) g (or more precisely, using exact calculation: \(795.5801\times159.808 = 795.5801\times(160 - 0.192)=795.5801\times160 - 795.5801\times0.192 = 127292.816 - 152.7513792 = 127140.0646\approx1.27\times10^{5}\) g. Wait, but let's do the calculation step by step with the given numbers:

\(n=\frac{5.76\times10^{4}}{72.4}=\frac{57600}{72.4}\). Let's compute 57600 ÷ 72.4:

72.4 × 795 = 72.4×800 - 72.4×5 = 57920 - 362 = 57558. Then 57600 - 57558 = 42, so 795 + 42/72.4 ≈795.5801 mol.

Then mass = 795.5801 mol × 159.808 g/mol. Let's compute 795.5801 × 159.808:

795.5801 × 159.808 = 795.5801 × (160 - 0.192) = 795.5801×160 - 795.5801×0.192 = 127292.816 - 152.7513792 = 127140.0646 g ≈1.27×10^{5} g (since 5.76 has 3 sig figs, 72.4 has 3, so the result should have 3 sig figs. Wait, 5.76×10^{4} is 3 sig figs, 72.4 is 3, so the ratio gives 3 sig figs, and molar mass of Br₂ is known precisely (we can consider it as exact for sig fig purposes here, or with more sig figs). So 127140 g is 1.27×10^{5} g when rounded to 3 sig figs. Alternatively, let's check the calculation again:

Wait, maybe I made a mistake in stoichiometry. The reaction is 2 HBr(g) + 72.4 kJ → H₂(g) + Br₂(l). So 72.4 kJ produces 1 mole of Br₂. So the moles of Br₂ is (5.76×10⁴ kJ) / (72.4 kJ/mol Br₂) = (5.76×10⁴ / 72.4) mol Br₂. Then mass is moles × molar mass of Br₂ (159.808 g/mol).

Calculating 5.76×10⁴ / 72.4:

5.76×10⁴ = 57600

57600 ÷ 72.4: Let's do this division more accurately. 72.4 × 795 = 72.4*(800 - 5) = 57920 - 362 = 57558. 57600 - 57558 = 42. So 42 ÷ 72.4 ≈0.5801. So total is 795.5801 mol.

Then 795.5801 mol × 159.808 g/mol:

795.5801 × 159.808:

Let's compute 795.5801 × 160 = 127292.816

Subtract 795.5801 × 0.192 = 795.5801 × 0.2 - 795.5801 × 0.008 = 159.11602 - 6.3646408 = 152.7513792

127292.816 - 152.7513792 = 127140.0646 g ≈1.27×10⁵ g (3 sig figs). Alternatively, if we use more precise calculation:

5.76×10⁴ = 57600

57600 ÷ 72.4 = 576000 ÷ 724 = let's divide 576000 by 724:

724 × 795 = 724*(800 - 5) = 579200 - 3620 = 575580

576000 - 575580 = 420

420 ÷ 724 ≈0.5801

So total is 795.5801 mol.

Then 795.5801 × 159.808 = 795.5801 × 159.808. Let's use a calculator approach:

159.808 × 700 = 111865.6

159.808 × 95 = 159.808×(100 - 5) = 15980.8 - 799.04 = 15181.76

159.808 × 0.5801 ≈159.808×0.5 = 79.904, 159.808×0.08 = 12.78464, 159.808×0.0001≈0.01598. So total ≈79.904 + 12.78464 + 0.01598≈92.70462

Now sum 111865.6 + 15181.76 = 127047.36 + 92.70462≈127140.0646 g, which is 1.27×10⁵ g (3 sig figs).

Wait, but maybe the correct answer is 1.27×10⁵ g, or let's check with exact calculation:

5.76×10⁴ kJ / 72.4 kJ per mol Br₂ = (5.76 / 72.4) × 10⁴ mol Br₂ = (0.079558) × 10⁴ mol = 795.58 mol Br₂.

Mass = 795.58 mol × 159.808 g/mol = 795.58 × 159.808.

795.58 × 160 = 127292.8

Subtract 795.58 × 0.192 = 795.58×0.2 - 795.58×0.008 = 159.116 - 6.36464 = 152.75136

127292.8 - 152.75136 = 127140.0486 g ≈1.27×10⁵ g (3 significant figures).

Yes, so the mass of Br₂ produced is approximately \(1.27\times10^{5}\) grams.

Answer:

\(1.27\times10^{5}\)