Nitroglycerine is an explosive used by the mining industry. It detonates according to the following equation: $$ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}(l) \longrightarrow 12 \mathrm{CO}_{2}(g)+6 \mathrm{~N}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g) $$ What volume is occupied by the gases produced when \(10.00 \mathrm{~g}\) of nitroglycerine explodes? 'The total pressure is \(1.45 \mathrm{~atm}\) at \(523^{\circ} \mathrm{C}\).

The molar mass of nitroglycerine (C₃H₅N₃O₉) is \(3(12.01 + 1.01)+(3 + 9)(16.00)+3(14.01)=227.09\mathrm{~g/mol}\). To find the number of moles of nitroglycerine, use the formula: `number of moles = mass / molar mass`. So, $$ n_{C_{3}H_{5}N_{3}O_{9}} = \frac{10.00\mathrm{~g}}{227.09\mathrm{~g/mol}} = 0.044\mathrm{~mol} $$

Now, use the balanced chemical equation to find the number of moles of each gas produced: For CO₂: $$ 12\mathrm{~moles~CO_{2}} \longleftrightarrow 4\mathrm{~moles~C_{3}H_{5}N_{3}O_{9}} $$ So, $$ n_{CO_{2}} = 0.044 \times \frac{12}{4} = 0.132\mathrm{~mol} $$ For N₂: $$ 6\mathrm{~moles~N_{2}} \longleftrightarrow 4\mathrm{~moles~C_{3}H_{5}N_{3}O_{9}} $$ So, $$ n_{N_{2}} = 0.044 \times \frac{6}{4} = 0.066\mathrm{~mol} $$ For H₂O: $$ 10\mathrm{~moles~H_{2}O} \longleftrightarrow 4\mathrm{~moles~C_{3}H_{5}N_{3}O_{9}} $$ So, $$ n_{H_{2}O} = 0.044 \times \frac{10}{4} = 0.110\mathrm{~mol} $$ For O₂: $$ 1\mathrm{~mole~O_{2}} \longleftrightarrow 4\mathrm{~moles~C_{3}H_{5}N_{3}O_{9}} $$ So, $$ n_{O_{2}}=0.044 \times \frac{1}{4} = 0.011\mathrm{~mol} $$

Add the moles of CO₂, N₂, H₂O, and O₂ to find the total number of moles: $$ n_\text{Total} = n_{CO_{2}} + n_{N_{2}} + n_{H_{2}O} + n_{O_{2}} = 0.132 + 0.066 + 0.110 + 0.011 = 0.319\mathrm{~mol} $$

To use the ideal gas law, we need the temperature in Kelvin. Add 273 to the given temperature in Celsius to convert: $$ T = 523^{\circ}\mathrm{C} + 273 = 796\mathrm{K} $$

Now, use the ideal gas law: \(PV = nRT\), with the ideal gas constant \(R=0.0821 \mathrm{L\cdot atm / mol\cdot K}\). Rearrange the equation to find the volume: $$ V = \frac{nRT}{P} = \frac{0.319 \cdot 0.0821 \cdot 796}{1.45} = 14.27\mathrm{~L} $$ So, the volume occupied by the gases produced when 10.00 grams of nitroglycerine explodes is 14.27 liters.