Question

Consider the unbalanced ethanol combustion reaction: C2H5OH + xO2 → 3H2O +yCO2 with an RQ value of ⅔. After balancing the reaction formula, calculate the accumulated amount of ethanol, meth (in g) used in T=2 hours of a steady-state combustion time and at an oxygen consumption rate of ṁo = 60 g/mn. Carbon molar mass is 12 g/mol

Answer 1

C2H5OH + xO2 → 3H2O + yCO2

RQ value = 2/3

RQ value = CO_{2 eliminated} / O_{2 consumed}

2/3 = y/x

y = 2 & x = 3

Hence, the balanced equation is:

C2H5OH + 3O2 → 3H2O + 2CO2

Rate of consumption of oxygen = 60 g/mn

Oxygen consumed in 2 hrs = 60 * 120

= 7200 g

Moles of oxygen consumed = 7200 / 32

= 225

1 mole C2H5OH consume 3 moles O2.

225 moles O2 will be consumed by = 225 / 3

= 75 moles of C2H5OH

Molar mass of C2H5OH = 46.06 g/mole

Mass of C2H5OH consumed = 75 * 46.06

= 3454.5 g