What vol of CO2 measured @ STP will be formed by the rxn of 1.3 mol of oxygen w/ 9.0x10^-1 mol CH3CH2OH?

Answer 1

First off, you need the balanced equation.

CH3CH2OH + 3 O2 ---> 2 CO2 + 3 H2O

Given the amount of reactants, your limiting reactant is oxygen (0.9 moles of ethanol would require 2.7 moles of oxygen to burn completely, and you only have 1.3), so the number of moles and thus the volume of CO2 that you can generate is based on the amount of oxygen. 3 moles of oxygen yields 2 moles of CO2, so the number of moles of CO2 is 0.87 moles.

Rearrange the Ideal Gas Law to get volume, as such, and solve for volume (remembering that STP is 1 atm and 0 C or 273K):

V = nRT/P = 0.87 moles * 0.08206 * 273K / 1 atm
V = 19.5 L

Answer 2

at STP, as a million mole of CO2 gas occupies 22.4L and molecular mass is 44g PV=nRT...(a million) n= mass/molecular mass(mm) now via (a million), P*mm= (mass/V) * R*T at STP, P=1atm and T=273 alright to that end a million* 40 4= d * 0.0821 * 273 -> d= a million.964 g/L approx. (as d= mass/V)

Answer 3

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