Quiz 2 - Chemical Reactions | Fundamentals of Chemical Reactions

A combination reaction involves multiple reactants combining to form a single product.

The balanced equation for the combustion of butane is 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O, so the coefficient of O₂​ is 13.

Decomposition reactions involve a single compound breaking down into two or more simpler substances, as in the breakdown of potassium chlorate into potassium chloride and oxygen gas.

1. Calculate the moles of carbon in CO₂:

• The molar mass of CO₂ is 44.01 g/mol.
• The mass of CO₂ produced is 0.733 g.
• The number of moles of CO₂ is $\frac{0.733 \mathrm{g}}{44.01 \mathrm{g}.{\mathrm{mol}}^{-1}}$ ≈ 0.0167 mol.
• Each mole of CO₂ contains one mole of carbon, so the moles of carbon in the hydrocarbon is 0.0167 mol.

2. Calculate the moles of hydrogen in H₂O:

• The molar mass of H₂O is 18.02 g/mol.
• The mass of H₂O produced is 0.300 g.
• The number of moles of H₂O is $\frac{0.300 \mathrm{g}}{18.02 \mathrm{g}.{\mathrm{mol}}^{-1}}$ ​≈ 0.0167 mol.
• Each mole of H₂O contains 2 moles of hydrogen, so the moles of hydrogen in the hydrocarbon is 2 × 0.0167 mol = 0.0334 mol

3. Determine the simplest whole-number ratio:

• The ratio of moles of carbon to hydrogen is 0.0167 mol C : 0.0334 mol H.
• Simplifying this ratio by dividing both values by the smaller number of moles (0.0167 mol), we get 1 mol C : 2 mol H.

1. Calculate the moles of carbon in CO₂:

• The molar mass of CO₂ is 44.01 g/mol.
• The mass of CO₂ produced is 15.4 g.
• The number of moles of CO₂ is $\frac{0.733 \mathrm{g}}{44.01 \mathrm{g}.{\mathrm{mol}}^{-1}}$ ≈ 0.0167 mol.
• Each mole of CO₂ contains one mole of carbon, so the moles of carbon in the hydrocarbon is 0.0167 mol.

2. Calculate the moles of hydrogen in H₂O:

• The molar mass of H₂O is 18.02 g/mol.
• The mass of H₂O produced is 0.300 g.
• The number of moles of H₂O is $\frac{0.300 \mathrm{g}}{18.02 \mathrm{g}.{\mathrm{mol}}^{-1}}$ ​≈ 0.0167 mol.
• Each mole of H₂O contains 2 moles of hydrogen, so the moles of hydrogen in the hydrocarbon is 2 × 0.0167 mol = 0.0334 mol

3. Determine the simplest whole-number ratio:

• The ratio of moles of carbon to hydrogen is 0.0167 mol C : 0.0334 mol H.
• Simplifying this ratio by dividing both values by the smaller number of moles (0.0167 mol), we get 1 mol C : 2 mol H.

The balanced equation is 4 Al + 3 O₂ → 2 Al₂O₃​, so the coefficient of O₂ is 3.

The balanced equation is 2 KMnO₄ + 16 HCl → 2 KCl + 2 MnCl₂ + 5 Cl₂ + 8 H₂O, so the coefficient of HCl is 16.

Assume 100 g of the compound:

• Nitrogen: $\frac{30.4 \mathrm{g}}{14.01 \mathrm{g}.{\mathrm{mol}}^{-1}}$ ​= 2.17 mol
• Oxygen: $\frac{69.6 \mathrm{g}}{16.00 \mathrm{g}.{\mathrm{mol}}^{-1}}$ ​= 4.35 mol

The empirical formula is NO₂​. The empirical formula mass is 46 g/mol. The molecular formula mass is 92 g/mol, so $\frac{92 \mathrm{g}.{\mathrm{mol}}^{-1}}{46 \mathrm{g}.{\mathrm{mol}}^{-1}}$ = 2. Therefore, the molecular formula is N₂O₄​.

The empirical formula mass of C₂H₃O is 43 g/mol. The molecular formula mass is 172 g/mol, so $\frac{172 \mathrm{g}.{\mathrm{mol}}^{-1}}{43 \mathrm{g}.{\mathrm{mol}}^{-1}}$ = 4. Therefore, the molecular formula is C₈H₁₂O₄​.

Assume 100 g of the compound:

• Carbon: $\frac{30.4 \mathrm{g}}{14.01 \mathrm{g}.{\mathrm{mol}}^{-1}}$ ​= 2.17 mol

• Nitrogen: $\frac{30.4 \mathrm{g}}{14.01 \mathrm{g}.{\mathrm{mol}}^{-1}}$ ​= 2.17 mol

• Oxygen: $\frac{69.6 \mathrm{g}}{16.00 \mathrm{g}.{\mathrm{mol}}^{-1}}$ ​= 4.35 mol

The empirical formula is NO₂​. The empirical formula mass is 46 g/mol. The molecular formula mass is 92 g/mol, so $\frac{92 \mathrm{g}.{\mathrm{mol}}^{-1}}{46 \mathrm{g}.{\mathrm{mol}}^{-1}}$ = 2. Therefore, the molecular formula is C₆H₁₂O₆.