1. Calculate the number of molecules in 0.75 mol of \( \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \). 2. Calculate the number of Oxygen atoms in a 0.350 mol of \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \). 3. Sodium carbonate has \( 2.41 \times 10^{24} \) molecules. Calculate the mass of sodium carbonate. 4. Determine the empirical formula of a compound composed of \( 62 \% \mathrm{C}, 5.20 \% \mathrm{H}, 12.1 \% \) N , and \( 20.6 \% \mathrm{O} \). 5. Combustion analysis of 3.58 mg ethyl butyrate produces, \( 7.12 \mathrm{mg} \mathrm{CO}_{2} \), and 3.58 mg \( \mathrm{H}_{2} \mathrm{O} \). Calculate the mass of oxygen in ethyl butyrate. 6. A mass of 5.915 g of a compound containing on \( \mathrm{C}, \mathrm{H} \) and O was burned in excess oxygen. The combustion reaction resulted in \( 11.942 \mathrm{~g} \mathrm{CO}_{2} \) and 0.749 g of \( \mathrm{H}_{2} \mathrm{O} \). Determine the empirical formula of the compound. 7. A compound \( \mathrm{C}_{3} \mathrm{H}_{8} \) weighing 4.54 g undergoes combustion according to the reaction: \( \mathrm{C}_{3} \mathrm{H}_{8}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \) a. Balance the given equation. b. How much in grams is \( \mathrm{CO}_{2} \) produced. 8. Balance the following equations: a. \( \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{~s}) \rightarrow \mathrm{PbO}+\mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \) b. \( \mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{~s}) \rightarrow \quad \mathrm{N}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad+\quad \mathrm{O}_{2}(\mathrm{~g}) \) 9. The following reaction was conducted in the lab, here is the balanced equation: \[ \mathrm{K}_{2} \mathrm{CrO}_{4}(\mathrm{aq})+\quad 2 \mathrm{AgNO}_{3}(\mathrm{aq}) \rightarrow \quad \mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{~s})+\quad 2 \mathrm{KNO}_{3}(\mathrm{aq}) \] A mass of \( 1.345 \mathrm{~g} \mathrm{~K}_{2} \mathrm{CrO}_{4} \) was reacted with \( 2.536 \mathrm{~g} \mathrm{AgNO}_{3} \). a. Find the limiting reactant.

Let's solve each of the problems step by step. ### 1. Calculate the number of molecules in 0.75 mol of \( \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \). **Known Condition:** - Moles of \( \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} = 0.75 \, \text{mol} \) - Avogadro's number \( N_A = 6.022 \times 10^{23} \, \text{molecules/mol} \) **Calculation:** \[ \text{Number of molecules} = \text{moles} \times N_A = 0.75 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \] ### 2. Calculate the number of Oxygen atoms in a 0.350 mol of \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \). **Known Condition:** - Moles of \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} = 0.350 \, \text{mol} \) - Each molecule of \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \) contains 6 Oxygen atoms. **Calculation:** \[ \text{Number of Oxygen atoms} = \text{moles} \times N_A \times 6 = 0.350 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \times 6 \] ### 3. Calculate the mass of sodium carbonate given \( 2.41 \times 10^{24} \) molecules. **Known Condition:** - Number of molecules = \( 2.41 \times 10^{24} \) - Molar mass of \( \mathrm{Na}_2\mathrm{CO}_3 \) (Sodium Carbonate) = \( 2 \times 22.99 + 12.01 + 3 \times 16.00 \) **Calculation:** 1. Calculate moles: \[ \text{moles} = \frac{\text{number of molecules}}{N_A} = \frac{2.41 \times 10^{24}}{6.022 \times 10^{23}} \] 2. Calculate mass: \[ \text{mass} = \text{moles} \times \text{molar mass} \] ### 4. Determine the empirical formula of a compound composed of \( 62 \% \mathrm{C}, 5.20 \% \mathrm{H}, 12.1 \% \mathrm{N}, \) and \( 20.6 \% \mathrm{O} \). **Known Condition:** - Percentages of each element. **Calculation:** 1. Assume 100 g of the compound: - \( 62 \, \text{g C} \) - \( 5.20 \, \text{g H} \) - \( 12.1 \, \text{g N} \) - \( 20.6 \, \text{g O} \) 2. Convert to moles: - \( n_C = \frac{62}{12.01} \) - \( n_H = \frac{5.20}{1.008} \) - \( n_N = \frac{12.1}{14.01} \) - \( n_O = \frac{20.6}{16.00} \) 3. Divide by the smallest number of moles to find the ratio. ### 5. Calculate the mass of oxygen in ethyl butyrate from combustion analysis. **Known Condition:** - Mass of \( \mathrm{CO}_{2} = 7.12 \, \text{mg} \) - Mass of \( \mathrm{H}_{2} \mathrm{O} = 3.58 \, \text{mg} \) **Calculation:** 1. Calculate moles of \( \mathrm{CO}_{2} \) and \( \mathrm{H}_{2} \mathrm{O} \): - \( n_{CO_2} = \frac{7.12 \, \text{mg}}{44.01 \, \text{g/mol}} \) - \( n_{H_2O} = \frac{3.58 \, \text{mg}}{18.02 \, \text{g/mol}} \) 2. Calculate moles of Carbon and Hydrogen: - \( C = n_{CO_2} \) - \( H = 2 \times n_{H_2O} \) 3. Calculate mass of Oxygen in ethyl butyrate. ### 6. Determine the empirical formula of a compound from combustion data. **Known Condition:** - Mass of compound = \( 5.915 \, \text{g} \) - Mass of \( \mathrm{CO}_{2} = 11.942 \, \text{g} \) - Mass of \( \mathrm{H}_{2} \mathrm{O} = 0.749 \, \text{g} \) **Calculation:** 1. Calculate moles of \( \mathrm{CO}_{2} \) and \( \mathrm{H}_{2} \mathrm{O} \). 2. Calculate moles of Carbon and Hydrogen. 3. Calculate mass of Oxygen in the compound. 4. Determine the empirical formula. ### 7. Combustion of \( \mathrm{C}_{3} \mathrm{H}_{8} \). **a. Balance the equation:** \[ \mathrm{C}_{3} \mathrm{H}_{8} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O} \] **b. Calculate grams of \( \mathrm{CO}_{2} \) produced:** 1. Calculate moles of \( \mathrm{C}_{3} \mathrm{H}_{8} \). 2. Use stoichiometry to find moles of \( \mathrm{CO}_{2} \). 3. Convert moles of \( \mathrm{CO}_{2} \) to grams. ### 8. Balance the following equations. **a. \( \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{~s}) \rightarrow \mathrm{PbO}+\mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \)** **b. \( \mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{~s}) \rightarrow \quad \mathrm{N}_{2}(\mathrm{~g})+\mathrm{H}_{2