Question

1 mole of a hydrocarbon of formula CnH2n was burned completely in oxygen producing carbon dioxide and water vapour only. It required 192g of oxygen.

Answer 1

The student's question deals with the combustion of a hydrocarbon in Chemistry to find the empirical formula based on the given oxygen used during the reaction.

Step-by-step explanation:

The student is dealing with a combustion reaction in Chemistry, where a hydrocarbon (CnH2n) reacts completely with oxygen to produce carbon dioxide and water vapour. To solve the problem, we must first write the balanced chemical equation for the combustion of the hydrocarbon. The general equation is CnH2n + (3n/2) O2 → n CO2 + n H2O. Since it is given that 192g of oxygen was consumed, we can calculate the amount in moles, as the molar mass of oxygen (O2) is 32 g/mol. Thus, the moles of oxygen used are 192 g / 32 g/mol = 6 moles. This implies 6 moles of O2 requires 2/3 * 6 = 4 moles of the hydrocarbon for complete combustion. For 1 mole of the hydrocarbon, we get a different amount of O2.

One mole of the gas at standard temperature and pressure (STP) occupies 22.414 liters. So, to compute the moles of gas produced, we can use this information if the volume and conditions are known. The complete combustion of a hydrocarbon with oxygen always producing CO2 and H2O helps in determining the amounts of products formed based on the stoichiometry of the reaction, from which we can work backward to find the empirical formula or molar ratio of the combustion elements.

Answer 2

The question addresses the complete combustion of a hydrocarbon to produce carbon dioxide and water vapor, requiring 192g of oxygen. The solution involves applying stoichiometry and molar ratios to determine the molecular formula of the hydrocarbon based on the oxygen consumed.

Step-by-step explanation:

The question involves determining the reaction of combustion of a hydrocarbon which is a Chemistry concept typically encountered at the high school level. By given information that 1 mole of a hydrocarbon of formula CnH2n requires 192g of oxygen for complete combustion to produce carbon dioxide and water, one can deduce the molecular formula of the hydrocarbon based on the stoichiometry of the reaction and the amount of oxygen consumed.

A similar concept is demonstrated in the combustion of n-heptane (C7H16), which reacts with oxygen to give carbon dioxide and water. A balanced equation for this reaction is essential to determining the stoichiometry, which in turn allows us to calculate the number of moles of reactants and products involved. For every mole of carbon in the hydrocarbon, there's a mole of CO2 produced, and for every two moles of hydrogen, there's a mole of H2O produced in the combustion reaction.

For example, the complete combustion equation for one of the simplest hydrocarbons, ethane (C2H6), would be represented as following:

2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)

In practice, to solve for the molecular formula of a hydrocarbon using the given data of oxygen consumption, you would apply concepts of molecular weights, molar ratios from the balanced equation, and the conservation of mass.