What volume of oxygen at s.t.p would react with carbon to form 4.40g of CO2 according to the following equation? C(s) + O2(g) → CO2(g) [O = 16; C =12; 1 mol...
What volume of oxygen at s.t.p would react with carbon to form 4.40g of CO2 according to the following equation? C(s) + O2(g) → CO2(g) [O = 16; C =12; 1 mole of a gas occupies 22.4dm3 at s.t.p]
Answer Details
The balanced chemical equation is C(s) + O2(g) → CO2(g). From the equation, we can see that 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide. The molar mass of CO2 is 12 + 2(16) = 44 g/mol. Given that 4.40 g of CO2 is formed, the number of moles of CO2 can be calculated as: number of moles = mass / molar mass = 4.40 g / 44 g/mol = 0.1 mol Since 1 mole of oxygen reacts with 1 mole of carbon to produce 1 mole of CO2, the number of moles of oxygen required is also 0.1 mol. Using the ideal gas equation PV = nRT, we can find the volume of oxygen required at standard temperature and pressure (s.t.p), where P = 1 atm, V = ?, n = 0.1 mol, R = 0.0821 L.atm/K.mol, and T = 273 K: V = nRT / P = 0.1 mol x 0.0821 L.atm/K.mol x 273 K / 1 atm = 2.24 L 1 mole of any gas occupies 22.4 L at s.t.p, so the volume of oxygen required is 2.24 L, which is.