Stoichometery Problem

Question

how many grams of carbon dioxide would result if we burn 19.2 liters of pure methane measured at 0.813 atm and 26 C?

Answer 1

Write a correctly balanced equation:

CH₄ + 2O₂ ==> CO₂ + 2H₂O ... balanced equation

From the balance equation we see that 1 mole CH₄ produces 2 moles CO₂. So, the first thing we need to do is to find out how many moles of CH₄ we have used:

PV = nRT (ideal gas law)

n = PV/RT = (0.813 atm)(19.2 L)/(0.0821 Latm/Kmol)(299K) = 0.6359 moles CH₄

Now that we have the moles of CH₄, we can calculate moles of CO₂ produced (using the stoichiometry in the balanced equation), and then convert moles CO₂ to grams CO₂, as follows:

0.6359 mol CH₄ x 1 mol CO₂/1 mol CH₄ x 44 g/mol CO₂ = 28.0 g CO₂ (to 3 significant figures)

Answer 2

CH₄ (g) + O₂ (g) --> CO₂ (g) + H₂O (l) Equation is balanced. Now identify what is given in the problem.

Volume (V_CH4) = 19.2 L Pressure (P) = 0.813 atm Temperature (T) = 26 ^oC, but need to convert to Kelvin, so 26 + 273= 299K Not given in the problem is the ideal gas constant (R) = 0.0821 Latm/molK

Now use Ideal Gas Law: PV = nRT and solve for n ==> PV/RT = n

n = (0.813 atm X 19.2 L) / (0.0821 Latm/molK X 299K) = 0.635 mol of CH₄

The molar mass of CO₂ is 44 g/mol (12 g from C + (2 X 16 g of O)) = 12 + 32 = 44

Now do the stoichiometry: mol CH₄ to mol CO₂ to g CO₂

0.635 mol CH₄ X (1mol CO₂ / 1 mol CH₄) X (44 g CO₂ / 1 mol CO₂) = 27.9 g CO₂ . is the answer.