Question

Assume that 1 mol of carbon dioxide is 22.4 l, on what volume (in l) of carbon dioxide will be produced when you burn 20.0 mol of acetylene gas?

Answer 1

To determine the volume of carbon dioxide produced from burning 20.0 moles of acetylene, we use the stoichiometry of the combustion reaction, which shows 2 moles of acetylene yield 4 moles of CO2. This results in 40.0 moles of CO2, which, at STP translates to 896.0 liters of CO2 when using the molar volume of 22.4 L/mol.

Step-by-step explanation:

To calculate the volume of carbon dioxide produced from the combustion of acetylene, we begin with the balanced chemical equation for the combustion of acetylene (C2H2):

2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)

From the stoichiometry of the balanced equation, we see that 2 moles of acetylene produces 4 moles of carbon dioxide. So, 1 mole of acetylene produces 2 moles of carbon dioxide. If we have 20.0 moles of acetylene, we multiply by the stoichiometric ratio to find the moles of carbon dioxide:

20.0 moles C2H2 × × (2 moles CO2 / 1 mole C2H2) = 40.0 moles CO2

Using the conversion factor that at STP, 1 mole of a gas occupies 22.4 liters, we can find the volume of carbon dioxide:

40.0 moles CO2 × 22.4 L/mol CO2 = 896.0 liters of CO2