Question

Consider the combustion reaction between 26.0mL of liquid methanol (density = 0.850 g/mL) and 13.0L of oxygen gas measured at STP. The products of the reaction are CO2 (g) and H20 (g). Calculate the volume of liquid H2O formed if the reaction goes to completion and you condense the water vapor.

Answer 1

So firstly, write the chemical equation.

CH3OH(l) + O2(g) --> CO2(g) + H2O(g) we must balance it:

2CH3OH(l) + 3O2(g) --> 2CO2(g) + 4H2O(g)

First we must find the limiting reactant (find the moles of each because in this case, finding moles will be more efficient than finding grams):

Convert 26 mL of Methane to find moles of methane.
26 mL CH3OH ( 0.850g / 1 mL ) ( 1 mole / 32g ) = 0.69 moles of CH3OH

To find the moles of O2, we can use the ideal gas law equation: PV=(0.08206)(n)(T) Plug it in and solve for n (the number of moles). Since its at STP, we know that Pressure: 1 atm, Temperature: 273 K, and Volume: 13 L because there are 13 L of Oxygen

(1)(13)=(0.08206)(n)(273) = 13 = 22.402n = 0.5803 moles of O2

Now find the limiting reactant and set both reactions to equal x moles of H2O because that's what we're trying to find.

0.69 moles of CH3OH ( 4 moles H2O / 2 mole CH3OH ) = 1.38 moles H2O

0.5803 moles of O2 ( 4 moles H2O / 3 mole O2 ) = .77 moles H2O

The limiting reactant is O2 because .77 moles is less than 1.38 moles. So your answer is....
.77 moles H2O