how many liters of CO2 and water vapor are produced by burning .3 liters of C2H2 (Acetylene) with .75 liters of O2? The balanced equation is:

2C2H2 + 5O2 = 2 H2O + 4CO2

To find out how many liters of CO2 and water vapor are produced by burning a certain amount of C2H2 and O2, we can use stoichiometry.

First, we need to determine the limiting reactant, which is the reactant that will be completely consumed and determines the amount of product formed. To do this, we compare the number of moles of C2H2 and O2 available.

From the balanced equation, we can see that the stoichiometric ratio between C2H2 and O2 is 2:5. Therefore, for the given amounts of C2H2 and O2:

moles of C2H2 = volume (in liters) of C2H2
moles of O2 = volume (in liters) of O2

moles of C2H2 = 0.3 liters of C2H2
moles of O2 = 0.75 liters of O2

To find the limiting reactant, we compare the ratios of moles available for both reactants:

moles of C2H2 / 2 = 0.3 / 2 = 0.15 moles
moles of O2 / 5 = 0.75 / 5 = 0.15 moles

Since both moles are equal, neither C2H2 nor O2 is limiting. Therefore, both will react completely.

Next, we can determine the number of moles of CO2 and H2O produced using the stoichiometric ratios from the balanced equation:

moles of H2O = moles of C2H2 × (2 mol H2O / 2 mol C2H2) = 0.15 moles
moles of CO2 = moles of C2H2 × (4 mol CO2 / 2 mol C2H2) = 0.30 moles

Finally, we can convert the moles of CO2 and H2O to volumes (in liters) using the ideal gas law:

PV = nRT

Assuming the reaction occurs at standard temperature and pressure (STP), where R = 0.0821 L·atm/(mol·K) and T = 273 K, we have:

V = nRT / P

where V is the volume, n is the number of moles, R is the gas constant, T is the temperature, and P is the pressure (assuming 1 atm).

For H2O:
V of H2O = (0.15 moles H2O) × (0.0821 L·atm/(mol·K)) × (273 K) / (1 atm) ≈ 3.77 liters

For CO2:
V of CO2 = (0.30 moles CO2) × (0.0821 L·atm/(mol·K)) × (273 K) / (1 atm) ≈ 7.54 liters

Therefore, burning 0.3 liters of C2H2 and 0.75 liters of O2 will produce approximately 3.77 liters of H2O and 7.54 liters of CO2.

All are gases therefore, you may use as a shortcut that the L is directly the same as mols.

0.3L C2H2 x (2 H2O/2 C2H2) = ?L H2O produced.
0.3L C2H2 x (4 mols CO2/2 mols C2H2) = ? L CO2 produced.

0.3 L C2H2 x (5 mols O2/2 mols C2H2) = ? mols O2 used.