Question

Consider the combustion reaction between 25.0mL of liquid methanol (density=0.850g/mL) and 12.5 L of oxygen gas measured at STP. The products of the reaction are CO2(g) and H2O(g). Calculate the volume of liquid H2O (density= 1g/mL) formed if the reaction goes to completion and you condense the water vapor. *assume that oxygen is an ideal gas*

Answer 1

V = 25 ml of ethanol

mass = D*V = 0.85*25 = 21.25 g

mol methanol = mass/MW = 21.25/32 = 0.6640

V = 12.5 L of O2 at STP

note that at STP; gases (all of them) occupy 22.4 L/mol

12.5/22.4 = 0.5580 mol of O2

the reaction

CH3OH + O2 = CO2 + H2O

balanced

CH3OH + 2O2 = CO2 + 2H2O

then

calculate mol of water produced

ratio is 1:2 with respect of methanol and O2

then

0.6640 mol methanol need 0.6640*2 = 1.328 mol of O2

0.5580 mol of O2 need 0.5580 /2 = 0.279 mol of methanol, then O2 is limiting

0.279 mol of O2 need 2:2 ratio with respect of H2O to form it

then

0.279 mol of O2 forms 0.279mol ofH2O

mass = molo*MW = 0.279*18 = 5.022 g of H2O

if D = 1 g/ml then

V = 5.022 ml of H2O is produced