8g of \( \mathrm{CH_4} \) occupies 11.2 at S.T.P. What volume would 22g of \( \mathrm{CH_3CH_2CH_3} \) occupy under the same condition?

The volume occupied by a certain amount of a gas at Standard Temperature and Pressure (S.T.P) can be calculated using the Ideal Gas Law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Since the temperature and pressure are constant at S.T.P, we can use the following formula to calculate the volume occupied by a certain amount of gas: V = nRT/P, where n is the number of moles of the gas. So, if 8g of CH4 occupy 11.2 dm3 at S.T.P, then we can calculate the number of moles of CH4 using the molar mass of CH4 (16 g/mol) as follows: n = 8g / 16 g/mol = 0.5 mol And the volume occupied by 0.5 mol of CH4 at S.T.P can be calculated as follows: V = nRT/P = 0.5 * 8.31 * 273 / 101.3 = 11.2 dm3 Using the same approach, we can calculate the volume occupied by 22g of CH3CH2CH3 at S.T.P as follows: n = 22g / (3 * 12 + 6 * 1 + 3 * 16) g/mol = 0.5 mol V = nRT/P = 0.5 * 8.31 * 273 / 101.3 = 11.2 dm3 Therefore, 22g of CH3CH2CH3 would occupy 11.2 dm3 at S.T.P.