Determine the mass of O2(g) that can be produced from the unbalanced reaction

Question

With steps, please

Determine the mass of O2(g) that can be produced from the unbalanced reaction:

Consider the mixing of 4.1 L of CO2(g) and 2.1 L of  H2O(g) at 33°C and 730 mm Hg.

CO2(g) + H2O(g) = C4H10(l) + O2(g)

Mass is what is g?

Answer 1

First, we must balance the reaction:

8CO₂(g) + 10H₂O(g) ==> 2C₄H₁₀(l) + 13O₂(g) ... balanced equation

Next, we will find the moles of both reactants, i.e. moles CO₂ and moles H₂O present. To do this, we will employ the ideal gas law PV = nRT and solve for moles, n = PV/RT

P = 730 mm Hg x 1 atm/760 mm Hg = 0.961 atm

T = 33 + 273 = 306K

R = 0.0821 Latm/Kmol

For CO₂: n = (0.961)(4.1)/(0.0821)(306) = 0.157 moles CO₂

For H₂O: n = (0.961)(2.1)/(0.0821)(306) = 0.080 moles H₂O

Since it takes 10 moles H₂O for each 8 moles CO₂, clearly H₂O is present in limiting supply and it alone will dictate how much O₂ can be formed.

Final step is to now convert moles of H₂O present to moles of O₂ formed, and then to grams of O₂ formed:

0.080 moles H₂O x 13 moles O₂ / 10 moles H₂O = 0.104 moles O₂ formed

0.104 moles O₂ x 32 g /mol = 3.3 g O₂ formed